Analysis 1 & 2 — Recap
Typeset recap for ETH ITET Analysis 1 and Analysis 2: definitions, when-to-use maps, worked examples, and exercises. Read straight through or jump to a chapter below.
Analysis 1 — core topics #
- Foundations — logic, maps, sup/inf, induction; the language for every proof later.
- Sequences — convergence, Cauchy sequences, and a limit triage map.
- Series — convergence tests, power series, and when each test fits.
- Functions — continuity, compactness, and the theorems behind existence proofs.
- Sequences of functions — pointwise vs uniform convergence; when swapping limits is safe.
- Derivatives — rules, mean value theorem, Taylor, and Newton.
- ODEs — first-order and linear systems; homogeneous + particular workflows.
- Differential calculus in $\mathbb{R}^n$ — partials, Jacobian, chain rule, implicit functions.
- Extrema — free and constrained optimization; Hessian and Lagrange.
- Integration (Ch. 10) — Riemann, FTC, techniques, line integrals, and solids of revolution.
Analysis 2 — core topics #
- Integration (Ch. 11) — Jacobians, Green, Gauss, and volume and flux in $\mathbb{R}^n$.
- Potential fields — gradients as potentials; path independence and holes.
- Vector analysis — grad, div, rot, and the Green–Stokes–Gauss ladder.
- Topology — open sets, compactness, simply connected domains.
- Reference formulas — algebra, trig, linear algebra, geometry, and $dV$ factors at a glance.
- Complex numbers — polar form, holomorphic functions, Cauchy, and residues.
- Additional integrals — substitution templates beyond the standard table.
- Fourier series — orthogonality, coefficients, convergence, Parseval.
Chapter 1 — Foundations (Grundlagen) #
Logic, sets, maps, bounds on $\mathbb{R}$, summation tools, induction, partial fractions, and inequalities — the language every later chapter uses.
Story of the chapter: before limits and derivatives you need precise statements (“for all”, “there exists”), correct negations, control of sup/inf on $\mathbb{R}$, and algebraic templates (telescoping sums, partial fractions) that reappear in Ch 2, Ch 3, and Ch 10.
Prerequisites: secondary-school algebra.
Foundations ladder (how the pieces fit)
| Step | Tool | Feeds into |
|---|---|---|
| 1 | Quantifiers + negation (§1.1) | $\varepsilon$–$N$ proofs, divergence (Ch 2) |
| 2 | Maps injective/surjective (§1.2) | Functions, inverses (Ch 4) |
| 3 | sup/inf + triangle inequality (§1.3, §1.4) | Monotone limits, estimates (Ch 2, Ch 8) |
| 4 | Telescoping + induction (§1.5, §1.6) | Series (Ch 3) |
| 5 | Partial fractions (§1.7) | $\int P/Q$ (Ch 10) |
Where to start
Proofs / negating limits? → §1.1.
Injective, surjective, inverse? → §1.2.
Bounds on $\mathbb{R}$ or estimates with $\lvert\cdot\rvert$? → §1.3–§1.4.
1.1 Quantifiers #
1.1.1 Overview #
Idea
Quantifiers tell you who must satisfy a property: everyone ($\forall$), someone ($\exists$), exactly one ($\exists!$), or nobody ($\nexists$).
Quantifiers
| Symbol | Meaning | Example |
|---|---|---|
| $\forall x$ | for all $x$ | $\forall x\in\mathbb{N}: x>-1$ |
| $\exists x$ | there exists at least one $x$ | $\exists x\in\mathbb{R}: x>1$ |
| $\exists! x$ | there exists exactly one $x$ | $\exists! x\in\mathbb{R}_{>0}: x^2=1$ |
| $\nexists x$ | there is no such $x$ | $\nexists x\in\mathbb{N}: x=1.5$ |
Logic connectives
| Symbol | Meaning |
|---|---|
| $A\land B$ | and |
| $A\lor B$ | or |
| $\neg A$ | not |
| $A\Rightarrow B$ | if $A$ then $B$ |
Sets (notation)
| Symbol | Meaning | Example |
|---|---|---|
| $x\in A$ | $x$ is in $A$ | $2\in\mathbb{N}$ |
| $A\subseteq B$ | every element of $A$ lies in $B$ | $\mathbb{Q}\subseteq\mathbb{R}$ |
| $A\cup B$, $A\cap B$ | union, intersection | $\{2,3\}\cap\{1,4\}=\varnothing$ |
| $\varnothing$ | empty set |
Typical mistake Confusing $\forall$ with $\exists$: “for all” is a strong claim (one counterexample kills it); “there exists” is weak (one witness is enough).
1.1.2 Negation rules #
Idea
To disprove “for all”, find one bad $x$. To disprove “there exists”, show every $x$ fails.
| Statement | Negation |
|---|---|
| $\neg(A\lor B)$ | $\neg A\land \neg B$ (De Morgan) |
| $\neg(A\land B)$ | $\neg A\lor \neg B$ (De Morgan) |
| $\neg(\forall x,\,A(x))$ | $\exists x,\,\neg A(x)$ |
| $\neg(\exists x,\,A(x))$ | $\forall x,\,\neg A(x)$ |
| $\neg(A\Rightarrow B)$ | $A\land \neg B$ |
| $A\Rightarrow B$ | contrapositive: $\neg B\Rightarrow \neg A$ |
When to use
- Rewriting proof goals (“show not convergent” $\Leftrightarrow$ “$\exists\varepsilon$ …”)
- Checking whether a statement is false
Proof moves (exam level): direct proof; contrapositive ($A\Rightarrow B$ via $\neg B\Rightarrow\neg A$); proof by contradiction (assume $\neg B$, derive absurdity).
Example: negate a limit-style claim
Claim: $\forall\varepsilon>0\,\exists N\in\mathbb{N}\,\forall n\ge N:\ |a_n-L|<\varepsilon$ (definition of $a_n\to L$).
Negation (divergent / not $L$): $\exists\varepsilon>0\,\forall N\in\mathbb{N}\,\exists n\ge N:\ |a_n-L|\ge\varepsilon$.
Check: existence and order of quantifiers flipped — that is always the pattern for $\neg(\forall\exists\forall)$.
1.2 Maps (Abbildungen) #
1.2.1 Surjectivity #
Idea
Surjective = every target $y$ is hit at least once; the map “covers” all of $B$.
$f:A\to B$ is surjective iff
When to use
- Showing $f$ hits every value in the codomain; constructing preimages
Typical mistake Checking only a few sample $y$ values instead of all of $B$.
1.2.2 Injectivity #
Idea
Injective = different inputs never give the same output; you can undo on the image.
$f:A\to B$ is injective iff
(equivalently: $x_1\ne x_2\Rightarrow f(x_1)\ne f(x_2)$).
Useful test (Ch 6): if $f$ is differentiable and $f'>0$ everywhere on an interval, then $f$ is strictly increasing and hence injective there.
1.2.3 Bijectivity and inverse #
$f$ is bijective iff it is injective and surjective.
Idea
Bijection = perfect pairing between $A$ and $B$.
If $f:A\to B$ is bijective, the inverse $f^{-1}:B\to A$ satisfies
Typical mistake Defining $f^{-1}$ when $f$ is not bijective (e.g. $e^x:\mathbb{R}\to\mathbb{R}$ — restrict codomain to $(0,\infty)$ first).
1.2.4 Composition example #
If $f:X\to Y$ and $g:Y\to X$ satisfy $g\circ f=\mathrm{id}_X$, then $f$ is injective and $g$ is surjective.
Example: $e^x$ and $\ln$ — restrict the codomain
Setup: $f:\mathbb{R}\to\mathbb{R}$, $f(x)=e^x$, and $g(y)=\ln y$ on $(0,\infty)$.
- $f$ is injective (strictly increasing) but not surjective onto $\mathbb{R}$.
- $g\circ f(x)=\ln(e^x)=x$ on $\mathbb{R}$, so $f$ is injective.
- $f\circ g(y)=e^{\ln y}=y$ on $(0,\infty)$: a true inverse only after restricting the target to $(0,\infty)$.
1.2.5 Composition rules #
For $f:A\to B$ and $g:B\to C$:
| If … | Then … |
|---|---|
| $f,g$ injective | $g\circ f$ injective |
| $f,g$ surjective | $g\circ f$ surjective |
| $f,g$ bijective | $g\circ f$ bijective, and $(g\circ f)^{-1}=f^{-1}\circ g^{-1}$ |
Typical mistake Assuming $g\circ f$ surjective when only $f$ is surjective (need $g$ surjective on all of $B$).
1.3 Supremum and infimum #
Idea
The supremum is the least upper bound: not necessarily a maximum (if the set is open), but nothing below it is an upper bound.
- Supremum $s=\sup A$: $a\le s$ for all $a\in A$, and if $o$ is any upper bound then $s\le o$.
- Infimum $i=\inf A$: $a\ge i$ for all $a\in A$, and if $m$ is any lower bound then $i\ge m$.
Key identities
Completeness of $\mathbb{R}$: every non-empty set bounded above has a supremum in $\mathbb{R}$.
Link to Ch 2: a monotone bounded sequence converges to its sup/inf.
$\max$ vs $\sup$: on a closed bounded set $K$, a continuous function attains $\max$ and $\min$ on $K$ (Ch 4 EVT); for open sets, only sup/inf need exist.
Typical mistake Assuming $\sup A\in A$ (e.g. $\sup(0,1)=1$ but $1\notin(0,1)$).
1.4 Triangle inequality #
For $x,y\in\mathbb{R}$ (and in $\mathbb{C}$ with modulus):
Use when
Typical mistake Writing $\lvert x+y\rvert=\lvert x\rvert+\lvert y\rvert$ without checking signs (equality in $\mathbb{R}$ only when $x,y\ge 0$ or both $\le 0$).
1.5 Manipulation of sums and products #
1.5.1 Telescoping sums #
Idea
Consecutive terms cancel in pairs, leaving only boundary pieces.
Index shift: $\displaystyle\sum_{k=m}^{n} a_k=\sum_{j=m}^{n} a_j$ — rename the index when aligning with a formula.
1.5.2 Product telescoping #
1.5.3 Standard finite sums (use or prove by §1.6) #
Link to Ch 3: telescoping and these sums appear in partial-sum arguments.
Example: telescoping
Goal: evaluate $\displaystyle\sum_{k=1}^{n}\frac{1}{k(k+1)}$.
Step: $\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$.
Check: $n=1$ gives $1/2$.
1.6 Complete induction #
Idea
Prove a statement for all integers $\ge n_0$ by (i) checking the first case, (ii) showing: if it holds for $n$, it holds for $n+1$ — like dominoes.
To prove $A(n)$ for all $n\ge n_0$, $n\in\mathbb{N}$:
- Base: verify $A(n_0)$.
- Inductive hypothesis: assume $A(n)$ for some $n\ge n_0$.
- Step: prove $A(n+1)$ using the hypothesis.
When to use
- Sums, divisibility, inequalities with $n$
Typical mistake Using $A(n+1)$ as if already proved without stating the inductive hypothesis; trying to induct a false universal statement (see Ex C1).
Example
Claim: $\displaystyle\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$.
Base $n=1$: $1=1$.
Step: assume true for $n$. Then
Check: matches the $\sum k$ formula in §1.5.3 with $n+1$ in place of $n$.
1.7 Partial fraction decomposition #
Idea
Write a rational function as simpler fractions — one per pole — so integration (Ch 10) and series become routine.
Workflow (exam recipe)
- If $\deg P\ge\deg Q$, perform polynomial division: $\dfrac{P}{Q}=S+\dfrac{R}{Q}$ with $\deg R<\deg Q$.
- Factor $Q$; find all zeros.
- Assign unknown constants to each partial template.
- Clear denominators or compare coefficients.
- Verify by recombining.
Real simple pole at $x_0$:
Real $r$-fold pole:
Irreducible quadratic $ax^2+bx+c$ with discriminant $b^2-4ac<0$ (no real roots):
Complex $r$-fold root: same template with linear numerators (coefficients may be complex; recombine to real fractions when possible).
Example
Goal: decompose $\dfrac{1}{x^2-1}$.
Coefficients: $A=\tfrac12$, $B=-\tfrac12$, so
Check: recombine to $\dfrac{x+1-(x-1)}{2(x^2-1)}=\dfrac{1}{x^2-1}$.
1.8 Method map (quick reference) #
| Task | Tool |
|---|---|
| Disprove “for all …” | Find one counterexample |
| Prove injectivity | $f(x_1)=f(x_2)\Rightarrow x_1=x_2$ |
| Compose maps | §1.2.5 |
| Bound a set on $\mathbb{R}$ | sup / inf (§1.3) |
| Simplify $\sum$ | Telescoping or §1.5.3 |
| Integrate rational $P/Q$ | §1.7 then Ch 10 |
| Prove for all $n$ | Induction (§1.6) |
1.9 Foundations — Exercises #
A) Direct application #
A1 Negate: $\forall x\in\mathbb{R}\,\exists y\in\mathbb{R}:\ x+y=0$.
A2 Is $f:\mathbb{R}\to\mathbb{R}$, $f(x)=x^3$ injective? surjective? bijective?
A3 Compute $\displaystyle\sum_{k=1}^{n}\frac{1}{4k(k+1)}$.
A4 Find $\sup$ and $\inf$ of $A=\{x\in\mathbb{Q}: x^2<2\}$ in $\mathbb{R}$.
B) Multi-step #
B1 Decompose $\dfrac{x^2+1}{x^3-x}$ into partial fractions.
B2 Prove by induction: $2^n\ge n+1$ for all $n\in\mathbb{N}$.
B3 Prove: if $f,g$ are injective, then $g\circ f$ is injective (§1.2.5).
C) Mixed exam-style #
C1 Let $A(n)$ be “$n^2+n+41$ is prime”. Test small $n$, then discuss why induction of a false universal claim fails (find $n$ with $A(n)$ false).
C2 Decompose $\dfrac{2x+3}{(x-1)^2(x+2)}$ and integrate (preview Ch 10).
Answers (A) #
A1 $\exists x\in\mathbb{R}\,\forall y\in\mathbb{R}:\ x+y\ne 0$.
A2 Injective yes ($x_1^3=x_2^3\Rightarrow x_1=x_2$); surjective yes (every $y$ has root $y^{1/3}$); bijective yes.
A3 $\dfrac{1}{4}\bigl(1-\dfrac{1}{n+1}\bigr)=\dfrac{n}{4(n+1)}$.
A4 $\sup A=\sqrt{2}$, $\inf A=-\sqrt{2}$ (neither lies in $A$).
Answers (B) #
B1 $\dfrac{x^2+1}{x^3-x}=\dfrac{-1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x+1}$ (see worked solution below).
B2 Base $n=1$: $2\ge2$. Step: $2^{n+1}=2\cdot 2^n\ge 2(n+1)=2n+2\ge n+2$ for $n\ge 1$.
B3 $g(f(x_1))=g(f(x_2))\Rightarrow f(x_1)=f(x_2)\Rightarrow x_1=x_2$ by injectivity of $f$ then $g$.
Answers (C) #
C1 $A(n)$ holds for many small $n$ (e.g. $n=1,\ldots,40$) but fails at $n=41$ since $41^2+41+41=41\cdot 43$ is not prime. Induction cannot prove a false universal statement.
C2 Same decomposition as the representative solution below; integrate in Ch 10.
Worked solution (A4 — sup not in the set) #
Set: $A=\{x\in\mathbb{Q}: x^2<2\}\subset\mathbb{R}$.
Every $a\in A$ satisfies $a<\sqrt{2}$ and $a>-\sqrt{2}$. The numbers $\pm\sqrt{2}$ are upper/lower bounds but not in $A$ (irrational).
- Upper bounds: any $o\ge\sqrt{2}$ works; the least upper bound is $\sup A=\sqrt{2}$.
- Lower bounds: symmetrically $\inf A=-\sqrt{2}$.
Check: for any $\varepsilon>0$, $\sqrt{2}-\varepsilon\in A$ (choose rational $q$ with $\sqrt{2}-\varepsilon<q<\sqrt{2}$), so no smaller number is an upper bound.
Worked solution (B1 — partial fractions) #
$\dfrac{x^2+1}{x^3-x}=\dfrac{x^2+1}{x(x-1)(x+1)}$ with $\deg$ numerator $< \deg$ denominator.
Ansatz $\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{x+1}$. Clear denominators:
$x=0\Rightarrow 1=-A\Rightarrow A=-1$. $x=1\Rightarrow 2=2B\Rightarrow B=1$. $x=-1\Rightarrow 2=2C\Rightarrow C=1$.
Check: recombine the right-hand side over $x(x-1)(x+1)$ to recover $x^2+1$.
Representative solution (C2) #
Partial fractions: $\dfrac{2x+3}{(x-1)^2(x+2)}=\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+2}$.
Solving: $A=-\tfrac{1}{9}$, $B=\tfrac{7}{9}$, $C=\tfrac{1}{9}$ (verify by recombination).
Quick decision checklist #
- Negate a statement? → flip quantifiers (§1.1.2).
- Map properties? → definitions in §1.2; composition in §1.2.5.
- Least upper bound / gap in set? → sup/inf (§1.3).
- Estimate $\lvert\cdot\rvert$? → triangle inequality (§1.4).
- Sum $\sum_{k=1}^n(\cdots)$? → telescope (§1.5.1) or known formula (§1.5.3).
- Claim for all $n\in\mathbb{N}$? → induction (§1.6) — first check the claim is true.
- Rational $P/Q$? → §1.7, then Ch 10.
Chapter 2 — Sequences (Folgen) #
Convergence, bounds, Cauchy sequences, and limit computation — the bridge from algebra to analysis.
Story of the chapter: a sequence is an ordered list $a_1,a_2,\ldots$. Analysis asks whether the list settles near a number $L$. Monotone bounded sequences always settle; Cauchy’s criterion lets you prove convergence without knowing $L$ first.
Prerequisites: Chapter 1 (sup/inf, triangle inequality, quantifier negation §1.1).
2.1 Theorems on sequences #
2.1.1 Monotone bounded sequences (Theorem 2.8.3) #
Idea
A sequence that only moves upward but stays below a ceiling must approach that ceiling (the supremum). Falling sequences approach the infimum.
- If $(a_n)$ is monotone increasing and bounded above, then $a_n\to\sup\{a_n:n\in\mathbb{N}\}$.
- If $(a_n)$ is monotone decreasing and bounded below, then $a_n\to\inf\{a_n:n\in\mathbb{N}\}$.
Typical mistake Assuming convergence without proving boundedness (e.g. $a_n=n$ is monotone but diverges).
2.1.2 Cauchy sequences #
$(a_n)$ is a Cauchy sequence iff
In $\mathbb{R}$ (and $\mathbb{C}$): $(a_n)$ converges $\Leftrightarrow$ $(a_n)$ is Cauchy.
Use when You want to prove convergence without guessing the limit.
2.1.3 Contracting sequences (Satz 2.8.10) #
If $0\le c<1$ and $|a_{n+2}-a_{n+1}|\le c|a_{n+1}-a_n|$ for all $n$, then $(a_n)$ converges.
Example (ZSF): $a_{n+1}=\tfrac12(a_n+\sqrt{c})$ with $c\ge 1$, $a_1=\sqrt{c}$ converges to $\sqrt{c}$.
2.1.4 Bolzano–Weierstrass #
Every bounded sequence in $\mathbb{C}$ has at least one accumulation point (subsequence converging to some limit).
Use when Extracting convergent subsequences on compact domains (links Ch 9).
2.1.5 Divergence to $\pm\infty$ #
$a_n\to+\infty$ iff $\forall T\in\mathbb{R}\,\exists N:\ a_n>T$ for all $n\ge N$.
Similarly $a_n\to-\infty$ iff $\forall T$, eventually $a_n<T$.
2.2 Limits #
2.2.1 Important limits (memorize) #
| Limit | Value |
|---|---|
| $\lim_{n\to\infty}n^{-s}$ | $0$ for $s>0$ |
| $\lim_{n\to\infty}q^n$ | $0$ if $ |
| $\lim_{n\to\infty}\sqrt[n]{a}$ | $1$ for $a>0$ |
| $\lim_{n\to\infty}n^{1/n}$ | $1$ |
| $\lim_{n\to\infty}n^k z^n$ | $0$ if $ |
| $\lim_{x\to 0}\dfrac{1-\cos x}{x^2}$ | $\tfrac12$ |
| $\lim_{x\to 0^+}\ln x$ | $-\infty$ |
| $\lim_{x\to 0}\dfrac{\ln(1+x)}{x}$ | $1$ |
| $\lim_{x\to 0}\dfrac{a^x-1}{x}$ | $\ln a$ |
| $\lim_{x\to a}\left(1+\frac{1}{x}\right)^x$ | $e$ (with appropriate $a$) |
| $\lim_{x\to 0}\dfrac{\sin x}{x}$ | $1$ |
2.2.2 L’Hôpital’s rule #
For $f,g$ differentiable near $x_0$:
when the original limit is an indeterminate form: $\frac00$, $\frac{\infty}{\infty}$, $0\cdot\infty$, $\infty-\infty$, $0^0$, $\infty^0$, $1^\infty$.
Before applying: rewrite $0\cdot\infty$ and $\infty-\infty$ into a quotient.
Example ($0\cdot\infty$)
$\lim_{x\to 0^+} x\ln x$: set $x\ln x=\dfrac{\ln x}{1/x}$ ($\infty/\infty$), L’Hôpital gives $\dfrac{1/x}{-1/x^2}=-x\to 0$.
2.2.3 Special tricks #
- Sandwich: bound $a_n$ between two sequences with the same limit.
- Stolz–Cesàro (exam occasionally): for $\frac{0}{0}$ type $\frac{a_n}{b_n}$ when $b_n\to\infty$.
- Series comparison preview: if terms look like $1/n^p$, compare to Ch 3.
2.3 Method map — limit triage #
- Plug in — if continuous at the point, done.
- Standard table (§2.2.1)?
- Algebra — factor, conjugate, rationalize.
- Sandwich — especially with trig.
- L’Hôpital — only after indeterminate form confirmed.
- Sequence monotone + bounded — find limit via sup/inf.
Typical mistake Using L’Hôpital when the limit is not indeterminate (differentiate incorrectly).
2.9 Sequences — Exercises #
A) Direct #
A1 $\lim_{n\to\infty}\dfrac{3n^2+1}{n^2-n}$.
A2 $\lim_{n\to\infty}\dfrac{\sin n}{n}$.
A3 Show $(a_n)=(1+1/n)^n$ is increasing and bounded above by $3$ (hence convergent; limit is $e$).
B) Multi-step #
B1 $\lim_{x\to 0}\dfrac{e^x-1-x}{x^2}$ (L’Hôpital twice).
B2 Prove $a_{n+1}=\sqrt{2+a_n}$, $a_1=1$ converges and find the limit.
C) Mixed #
C1 $\lim_{n\to\infty}\sqrt{n^2+n}-n$ (hint: multiply by conjugate).
Answers #
A1 $3$. A2 $0$ (sandwich). A3 Standard $e$ proof outline.
B1 $\tfrac12$. B2 Limit $L$ satisfies $L=\sqrt{2+L}\Rightarrow L=2$.
C1 $\dfrac12$.
Chapter 3 — Series (Reihen) #
Infinite sums, convergence tests, and power series — sequences summed into a single number.
Story of the chapter: a series $\sum a_n$ is the limit of partial sums. Absolute convergence is the robust notion (rearrangements safe). Tests tell you which tool fits a given term shape.
Prerequisites: Chapter 2, §1.5 (telescoping sums).
3.1 Theorems on series #
3.1.1 Absolute convergence (Theorem 2.10.7) #
If $\sum_{n=1}^\infty |a_n|$ converges, then $\sum_{n=1}^\infty a_n$ converges and
3.1.2 Leibniz criterion #
If $(a_n)$ is monotone decreasing with $a_n\to 0$, then
converges (conditionally if $\sum|a_n|$ diverges).
3.1.3 Ratio test #
If $a_n\ne 0$ and $|a_{n+1}/a_n|\le c<1$ for all large $n$, then $\sum a_n$ converges absolutely.
If limit $=1$: test inconclusive.
3.1.4 Abel / partial summation (Serie 5 in ZSF) #
If $(a_n)$ decreases to $0$ and partial sums of $(b_n)$ are bounded, then $\sum a_n b_n$ converges.
3.1.5 Root test #
Let $q=\limsup_{n\to\infty}\sqrt[n]{|a_n|}$.
- $q<1$: absolute convergence
- $q>1$: divergence
3.2 Convergence criteria (decision tree) #
Idea
First check necessary condition $a_n\to 0$. Then match term shape to a test.
| Test | When to use |
|---|---|
| $p$-series $\sum 1/n^p$ | terms $\sim 1/n^p$ |
| Geometric $\sum r^n$ | $ |
| Ratio | factorials, $n!$, $x^n/n!$ |
| Root | $n$-th powers $(\cdots)^n$ |
| Comparison / limit comparison | rational-like terms |
| Integral test | $a_n=f(n)$ with $f$ positive, decreasing |
| Leibniz | alternating, decreasing to $0$ |
Standard facts
3.2.2 Necessary condition #
If $\sum a_n$ converges, then $a_n\to 0$. Contrapositive: if $a_n\not\to 0$, the series diverges.
3.2.3–3.2.4 Majorant / minorant #
Compare $|a_n|$ to a known convergent majorant, or compare positive terms to a divergent minorant.
3.2.5 Integral test #
If $f:[1,\infty)\to\mathbb{R}$ is positive, continuous, decreasing, then $\sum_{n=1}^\infty f(n)$ and $\int_1^\infty f(x)\,dx$ converge or diverge together.
3.3 Power series #
3.3.1 Definition #
is a power series about $x_0$.
3.3.2 Radius of convergence #
(or use ratio test between consecutive coefficients). Converges absolutely for $|x-x_0|<R$; diverges for $|x-x_0|>R$.
Inside the open disk: differentiate and integrate termwise (uniformly on compact subdisks).
3.4 Finite sum formulas #
3.5 Common infinite series (Taylor building blocks) #
Memorize: $e^x$, $\sin x$, $\cos x$, $\ln(1+x)$, $(1+x)^\alpha$ expansions — used heavily in Ch 6.
Link: full Fourier series on $[-\pi,\pi]$ in Chapter 18.
3.9 Series — Exercises #
A1 Does $\sum \frac{1}{n^{3/2}}$ converge?
A2 $\sum \frac{n!}{3^n}$ (ratio test).
A3 $\sum (-1)^{n+1}/\ln n$ (Leibniz).
B1 Find radius of convergence of $\sum \frac{x^n}{n^2}$.
C1 Decide $\sum \frac{\sin n}{n}$ (Abel / Dirichlet — bounded partial sums of $\sin$).
Answers #
A1 Yes ($p>1$). A2 Diverges (ratio $\to\infty$). A3 Converges conditionally.
B1 $R=1$. C1 Converges (Dirichlet).
Chapter 4 — Functions (Funktionen) #
Continuity, compactness consequences, and the analytic toolkit for limits of functions.
Story of the chapter: a function is continuous if small input changes produce small output changes. On closed bounded intervals, continuous functions attain maxima and minima and take all intermediate values — the theorems behind existence proofs everywhere.
Prerequisites: Chapters 2–3.
4.1 Theorems on functions #
4.1.1 Continuous functions #
$f:D\to\mathbb{R}$ is continuous at $x_0$ iff
Sequential characterization: $f$ continuous at $x_0$ $\Leftrightarrow$ for every sequence $x_n\to x_0$ in $D$, we have $f(x_n)\to f(x_0)$.
4.1.2–4.1.3 Lipschitz and uniform continuity #
Lipschitz: $|f(x)-f(y)|\le L|x-y|$ for some $L\ge 0$.
Uniform continuity: same $\delta$ works for all $x_0$ on the domain.
Lipschitz $\Rightarrow$ uniform continuity $\Rightarrow$ continuity (on bounded domains, not always equivalent).
4.1.5 Intermediate value theorem (IVT) #
If $f:[a,b]\to\mathbb{R}$ is continuous and $f(a)<c<f(b)$ (or reversed), then $\exists\xi\in(a,b): f(\xi)=c$.
4.1.6 Extreme value theorem #
Continuous $f$ on compact $[a,b]$ (or compact $K\subset\mathbb{R}^n$) attains a minimum and maximum on $K$.
4.1.7–4.1.8 Monotonicity and inverse #
Strictly monotone continuous $f$ on an interval has a continuous inverse $f^{-1}$ on $f(I)$.
4.1.9 Mean value theorem (preview) #
See Chapter 6: $\exists c$ with $f'(c)=\dfrac{f(b)-f(a)}{b-a}$.
4.2 Checking continuity #
4.2.1 One dimension #
- Elementary functions are continuous on their domain.
- Sums, products, quotients (where defined), compositions preserve continuity.
- For piecewise definitions: check limits agree at breakpoints.
4.2.2 Several dimensions #
$f:\Omega\subset\mathbb{R}^n\to\mathbb{R}^m$ is continuous at $x_0$ iff
Often proved via sequential characterization or Lipschitz bound on a neighborhood.
4.9 Exercises #
A1 Where is $f(x)=\dfrac{|x|}{x}$ continuous?
B1 Prove $f(x)=\sqrt{x}$ is uniformly continuous on $[0,1]$ but not Lipschitz on $[0,1]$.
C1 Show $x^3-3x+1=0$ has three real roots (IVT on intervals).
Chapter 5 — Sequences of functions (Funktionsfolgen) #
Pointwise vs uniform convergence — when you may swap limit with integral or derivative.
Story of the chapter: $f_n\to f$ pointwise if each fixed $x$ is hit eventually. Uniform convergence adds one speed $N(\varepsilon)$ for all $x$ at once — that extra strength is what makes limit theorems safe.
Prerequisites: Chapters 2–4.
5.1 Convergence modes #
5.1.1 Pointwise convergence #
$f_n:D\to\mathbb{R}$ converges pointwise to $f$ iff $\forall x\in D:\ f_n(x)\to f(x)$.
5.1.2 Uniform convergence #
Equivalent: $\|f_n-f\|_\infty\to 0$ on $D$.
Typical mistake Confusing pointwise with uniform — classic: $f_n(x)=x^n$ on $[0,1]$ converges pointwise but not uniformly.
5.1.3–5.1.5 Uniform limit of continuous functions #
If $f_n$ continuous and $f_n\to f$ uniformly on $D$, then $f$ is continuous.
Cauchy criterion for functions: $(f_n)$ uniformly Cauchy $\Leftrightarrow$ uniformly convergent (on complete codomain).
5.1.5 Normal convergence (series of functions) #
$\sum u_n$ converges normally if $\sum \|u_n\|_\infty<\infty$ $\Rightarrow$ uniform convergence of partial sums.
5.2 Recipe: prove uniform convergence #
- Estimate $\sup_{x\in D}|f_n(x)-f(x)|$.
- Show this supremum $\to 0$.
- Or use Weierstrass $M$-test for series: $|u_n(x)|\le M_n$, $\sum M_n<\infty$.
5.3 Swapping limit with derivative / integral #
| Operation | Need |
|---|---|
| $\displaystyle\lim_{n\to\infty}\int f_n=\int\lim f_n$ | uniform convergence (or dominated convergence — advanced) |
| $\displaystyle\frac{d}{dx}\lim f_n=\lim\frac{d}{dx}f_n$ | uniform convergence of derivatives (or uniform conv. + $f_n'$ continuous) |
Link: improper swapping in §10.1.4.
5.9 Exercises #
A1 Does $f_n(x)=\dfrac{x}{n}$ converge uniformly on $[0,1]$?
B1 $f_n(x)=x^n$ on $[0,1]$: find pointwise limit; uniform?
C1 Use $M$-test on $\sum \dfrac{\sin(nx)}{n^2}$ on $\mathbb{R}$.
Chapter 6 — Derivatives (Ableitungen) #
Rules, mean value theorem, convexity, Taylor polynomials, and Newton’s method.
Story of the chapter: the derivative is the best linear approximation of a function near a point. MVT turns average slope into an instantaneous slope somewhere inside; Taylor refines that to higher order.
Prerequisites: Chapters 4–5, limits from Chapter 2.
6.1 Basics and rules #
Product, quotient, chain rules; derivatives of inverse functions:
6.2 Common derivatives #
Polynomials, exponentials, logs, trig, inverse trig, hyperbolic and inverse hyperbolic (see ZSF table). Link: integral table in Chapter 10 §10.3.1.
6.4 Convex functions #
$f$ convex on $I$ iff $f''(x)\ge 0$ (if twice differentiable) or chord lies above graph.
Inequalities: Jensen-type bounds for exams; tangent line underestimates convex $f$.
6.5 Taylor polynomials #
Integral remainder (Lagrange):
for some $\xi$ between $a$ and $x$.
Product of series: build $f=gh$ from known Taylor factors (ZSF 6.5.3).
6.2.1 Newton’s method #
finds roots when started near a simple zero and $f'$ well-behaved.
6.9 Exercises #
A1 Derivative of $x^x$ ($x>0$).
B1 Taylor of $e^x$ at $0$ to order $3$ with remainder bound on $[-1,1]$.
C1 One Newton step for $f(x)=x^2-2$ from $x_0=1$.
Chapter 7 — Ordinary differential equations (ODEs) #
First-order and linear systems — modeling change and solving with structure.
Story of the chapter: an ODE relates a function to its derivatives. Linearity and constant coefficients let you split into homogeneous + particular parts; systems $\mathbf{x}'=A\mathbf{x}$ use eigenvalues of $A$.
Prerequisites: Chapter 6, Chapter 8 (matrix exponential, optional first pass).
7.1 General workflow #
- Classify: order, linear/nonlinear, homogeneous/inhomogeneous.
- Homogeneous solution $y_h$.
- Particular solution $y_p$ (ansatz, variation of constants, undetermined coefficients).
- General: $y=y_h+y_p$.
- Initial conditions fix free constants ($n$ conditions for $n$-th order).
7.2 Homogeneous linear ODEs #
First order: $y'+p(x)y=0$ $\Rightarrow$ $y=Ce^{-\int p}$.
Second order constant coefficients: $ay''+by'+cy=0$. Characteristic equation $ar^2+br+c=0$.
- Distinct real roots $r_1,r_2$: $C_1 e^{r_1 x}+C_2 e^{r_2 x}$.
- Double root $r$: $(C_1+C_2 x)e^{rx}$.
- Complex $\alpha\pm i\beta$: $e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)$ (Euler form: §16.1).
7.3 Particular solutions #
Inhomogeneous $ay''+by'+cy=f(x)$:
- Polynomial / exponential / sine ansatz when $f$ is a finite sum of such terms (adjust if resonance with homogeneous root).
- Variation of parameters when ansatz fails.
Separable: $y'=g(x)h(y)$ $\Rightarrow$ $\int dy/h(y)=\int g(x)\,dx$.
7.4 Matrix exponential #
For diagonalizable $A=PDP^{-1}$: $e^{At}=Pe^{Dt}P^{-1}$.
7.5 Systems #
Convert $n$-th order scalar ODE to first-order system in $\mathbb{R}^n$. Uniqueness: need $n$ independent initial conditions for an $n$-th order equation.
Link: $e^{At}$ and linear maps in Chapter 8.
7.9 Exercises #
A1 Solve $y'=2xy$, $y(0)=1$.
B1 $y''-3y'+2y=e^x$.
C1 System $\mathbf{x}'=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\mathbf{x}$ — periodic solutions (rotation).
Chapter 8 — Differential calculus in $\mathbb{R}^n$ #
Partial derivatives, total differential, Jacobian, chain rule, and implicit functions.
Story of the chapter: in several variables, derivative information lives in the gradient (vector of partials) and Jacobian (matrix). Differentiability means a good linear approximation; the inverse and implicit function theorems say when you can solve locally.
Prerequisites: Chapters 4, 6.
8.1 Concepts #
8.1.1 Partial derivative #
8.1.2 Directional derivative #
(if $f$ differentiable).
8.1.3 Total differentiability #
$f$ differentiable at $a$ iff
where $df(a):\mathbb{R}^n\to\mathbb{R}^m$ is linear (Jacobian matrix).
Hierarchy: $C^1$ $\Rightarrow$ differentiable $\Rightarrow$ partials exist; converse needs continuous partials.
8.1.5–8.1.6 Jacobian and gradient #
8.2 Theorems #
8.2.3 Schwarz #
If $f\in C^2$, then $\dfrac{\partial^2 f}{\partial x_i\partial x_j}=\dfrac{\partial^2 f}{\partial x_j\partial x_i}$.
8.2.4 Inverse function theorem #
If $f\in C^1$ and $df(x_0)$ invertible, then locally $f$ is a $C^1$ diffeomorphism onto its image.
8.2.6 Diffeomorphism checklist #
- $C^1$ and $\det df(x)\ne 0$ on $U$ $\Rightarrow$ local diffeomorphism.
- Bijection $U\to V$ $\Rightarrow$ global diffeomorphism (if constructed).
8.2.7 Implicit function theorem #
If $f(x,y)=0$ and $\det D_y f(a,b)\ne 0$, then locally $y=h(x)$ with $f(x,h(x))=0$ and
8.2.8 Existence of extrema on compact sets #
Continuous $f$ on compact $\Omega\subset\mathbb{R}^n$ attains min and max — used in Chapter 9.
8.9 Exercises #
A1 Compute $\nabla f$ for $f(x,y)=x^2y+y^3$.
B1 Implicit derivative: $x^2+y^2=1$, find $dy/dx$.
C1 Verify Schwarz on $f(x,y)=xy^2$.
Chapter 9 — Extrema (Extremwerte) #
Free and constrained optimization in $\mathbb{R}^n$.
Story of the chapter: interior extrema occur where the gradient vanishes (if differentiable). Whether a critical point is min/max/saddle is decided by the Hessian. Constraints add Lagrange multipliers.
Prerequisites: Chapter 8.
9.1 Free extrema #
Critical points: $\nabla f(a)=0$ (or $df(a)=0$).
Hessian test (2D, $C^2$): let $H=\det\begin{pmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{pmatrix}$ at $a$.
- $H>0$, $f_{xx}>0$: local minimum
- $H>0$, $f_{xx}<0$: local maximum
- $H<0$: saddle
- $H=0$: inconclusive
For $n\times n$ symmetric Hessians: eigenvalue test and Sylvester leading minors — §15.7.
Compact domain: compare critical points in interior with boundary values (EVT, Ch 8.2.8).
9.2 Lagrange multipliers #
Optimize $f(x)$ subject to $g(x)=0$. At a regular extremum:
Recipe: form $L=f-\lambda g$, set $\nabla L=0$ and constraint.
9.9 Exercises #
A1 Find and classify critical points of $f(x,y)=x^2+y^2-2x$.
B1 Maximize $xy$ on $x+y=1$, $x,y>0$.
C1 Closest point on $x^2+y^2=1$ to $(2,0)$.
Chapter 10 — Integration #
Limits, definite and improper integrals, integration techniques (substitution, parts, partial fractions, trig substitutions), line integrals, and solids of revolution — with exam-style practice at the end of the chapter.
Story of the chapter: we first define the integral (step functions, Riemann sums), then connect it to antiderivatives (FTC), then learn computational tools (§10.2–10.3) for integrals that are not elementary by inspection.
Prerequisites: limits and continuity (Ch 4), derivatives and FTC intuition (Ch 6), partial fractions (§1.7), uniform convergence when swapping limits (Ch 5 §5.3).
10.1 Definitions and Foundations #
10.1.1 Step Functions #
Idea
A step function is piecewise constant: on each subinterval it has one fixed value.
Definition Let $D = [a,b]$ with $a < b$ . A function $s : D \to \mathbb{R}$ is a step function if there is a partition
such that $s(t) = \sigma_i$ on each $(x_i, x_{i+1})$ .
Integral of a step function
Typical mistake Forgetting this is a signed area (negative values contribute negatively).
Example:
Goal: integrate a step function on $[0,4]$.
Setup: $s(t)=2$ on $[0,1)$, $s(t)=-1$ on $[1,3)$, $s(t)=3$ on $[3,4]$.
Check: signed area — positive pieces minus the negative piece on $[1,3)$.
10.1.2 Fundamental Theorem of Calculus (FTC) #
Idea
Think of $F(x)=\int_a^x g(t)\,dt$ as area accumulated from $a$ to $x$. Part I says that area changes at rate $g(x)$ (local height). Part II says total area over $[a,b]$ equals the net change of any antiderivative $G$ with $G'=g$.
Part A (derivative of integral)
if $g$ is continuous.
Part B (evaluation of definite integral)
where $G'(x)=g(x)$ .
Typical mistake Confusing the dummy variable $t$ inside the integral with the outside variable $x$ .
When not to use Part A If $g$ is not continuous on the interval you differentiate across (then $F'$ may fail at jumps).
Example:
Goal: use FTC in both directions for $g(t)=t^3+\cos t$.
Part A: $F(x)=\int_0^x (t^3+\cos t)\,dt$ gives $F'(x)=x^3+\cos x$ and $F(0)=0$.
Part B: an antiderivative is $G(x)=\frac{x^4}{4}+\sin x$, so
Check: Part A gives the rate; Part B gives the total — same underlying function $g$.
10.1.3 Riemann Sums #
Idea
An integral is the limit of sums of thin rectangle areas.
Here:
- $\Delta x = \frac{b-a}{n}$
- sample point $x_k = a + k\Delta x$ (left endpoints in the formula above; your sum may use a shifted index — that is fine if $x_k$ is consistent)
Identification checklist (do this before integrating)
- Rewrite the sum as $\displaystyle\sum_{k} f(x_k)\,\Delta x$.
- Read off $f$, $\Delta x$, and sample points $x_k$.
- Set $a = x_0$ (or the value at $k=0$ if your indexing starts there).
- Sanity check: $b-a = n\Delta x$ must hold. If not, $a$ or $b$ is wrong — stop and re-identify.
When to use Exam limits of the form $\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n (\cdots)$ or $\lim_{n\to\infty}\frac{c}{n}\sum (\cdots)$.
When not to use If you cannot match $b-a = n\Delta x$, or the sum is not a Riemann-type sample sum (e.g. missing $\Delta x$, or $x_k$ not affine in $k$).
Typical mistake Missing the factor $\Delta x$ , or forcing $a=0$ when the sample points show a shift (e.g. $1+\frac{2k}{n}$ means $a=1$, not $0$).
Example:
Goal: evaluate $\displaystyle\lim_{n\to\infty}\frac{2}{n}\sum_{k=1}^n\left(1+\frac{2k}{n}\right)^2$.
Identify: $\Delta x=\frac{2}{n}$, $x_k=1+\frac{2k}{n}$, so $f(x)=x^2$, $a=1$, $b=3$ (since $b-a=2=n\Delta x$).
Check: $b-a=2$ and $n\Delta x=2$ ✓; integrand $x^2$ matches $(1+2k/n)^2$ with $x_k=1+2k/n$.
10.1.4 Swapping Limit and Integral #
Idea
You want to pass the limit inside the integral.
If $f_n \to f$ uniformly on $[a,b]$ , then
Required hypothesis Uniform convergence of $f_n$ to $f$ on $[a,b]$.
When not to use Pointwise convergence alone — there exist sequences with $f_n\to f$ pointwise but $\int f_n \not\to \int f$.
Typical mistake Applying this under only pointwise convergence.
Example:
Let $f_n(x)=x+\frac{\sin(nx)}{n}$ on $[0,1]$ . Pointwise, $f_n(x)\to x$ . Also
so convergence is uniform. Therefore:
10.1.5 Improper Integrals #
Idea
The domain is too large or the integrand blows up — so you define the integral as a limit of proper integrals first, then ask whether that limit exists (converges).
Type 1: Infinite interval
Type 2: Endpoint singularity
(if $g$ is not defined or unbounded at $a$ ).
Two-sided infinite interval
and both integrals must converge.
Typical mistake Treating $\int_{-\infty}^{\infty}$ as a single symmetric limit without checking one-sided convergence.
Example:
Goal: decide convergence of $\displaystyle\int_1^\infty x^{-p}\,dx$ (the $p$-test).
For $p\neq 1$: $\displaystyle\int_1^R x^{-p}\,dx = \frac{R^{1-p}-1}{1-p}$. As $R\to\infty$:
- converges iff $p>1$, with value $\dfrac{1}{p-1}$;
- diverges for $p\le 1$.
Endpoint singularity: $\displaystyle\int_0^1 x^{-1/2}\,dx = \lim_{\varepsilon\downarrow 0}\int_\varepsilon^1 x^{-1/2}\,dx = 2$ (integrand blows up at $0$, but the limit exists).
Check: power $-p$ on an infinite interval — compare mentally to $\int_1^\infty \frac{1}{x}\,dx$ (divergent borderline).
10.1.6 Inequalities with Integrals #
If $g_1(x)\le g_2(x)$ on $[a,b]$ , then
If $g(x)\ge 0$ , then
If $g\ge 0$ and $[c,d]\subseteq [a,b]$ , then
Typical use Bounding hard integrals by easier comparison functions.
Example:
For $x\in[0,1]$ :
Integrate all sides:
so
This is a clean comparison bound.
10.1.7 Absolute Value Estimates #
if $|g(t)|\le M$ on the interval.
Interpretation Integral size is controlled by maximum height times interval length.
Typical mistake Using an $M$ that is not valid over the full interval.
Example:
Bound
Since $|\sin(t^2)|\le1$ :
Useful when exact antiderivative is unavailable.
10.1.8 Line Integrals (Wegintegrale) #
Idea
You integrate along a curve, not just on an interval. Scalar form: total mass of a wire with density $f$ along the path. Vector form: work $=\int \mathbf{F}\cdot d\mathbf{r}$ (force along displacement).
Scalar field along a curve (arc length form) #
For $f:\mathbb{R}^n\to\mathbb{R}$ and $\gamma:[a,b]\to\mathbb{R}^n$ :
Vector field work integral #
For $\mathbf{v}:\mathbb{R}^n\to\mathbb{R}^n$ and $\gamma:[a,b]\to\mathbb{R}^n$ :
Typical mistake Forgetting $\|\gamma'(t)\|$ in $ds$ -integrals.
Example:
Use the same path for both cases: $\gamma(t)=(t,t^2)$ , $t\in[0,1]$ . Then $\gamma'(t)=(1,2t)$ and $\|\gamma'(t)\|=\sqrt{1+4t^2}$ . Scalar case: let $f(x,y)=x$ . Then $f(\gamma(t))=t$ and
Vector case: let $\mathbf v(x,y)=(y,x)$ . Then $\mathbf v(\gamma(t))=(t^2,t)$ and the work integral is
10.1.9 Conservative Vector Fields #
Idea
You may use $\phi(B)-\phi(A)$ only when $\mathbf{v}=\nabla\phi$ on the whole region containing your path (and the region is simply connected — no holes; see §14.6). You do not get to guess $\phi$; you must check that its partial derivatives match $\mathbf{v}$.
Definition Write $\mathbf{v}(x,y)=(P(x,y),Q(x,y))$. The field is conservative on a domain $D$ if there is a potential $\phi$ on $D$ with
When you are allowed to use $\phi(B)-\phi(A)$
- You have found $\phi$ with $\nabla\phi=\mathbf{v}$ on $D$ (verified by differentiation).
- Your path from $A$ to $B$ lies entirely in $D$.
- $D$ is simply connected (or you stay in one branch of $\phi$).
When not to use
- You only “think” a formula looks like an antiderivative — you must check $\partial_x\phi=P$ and $\partial_y\phi=Q$.
- $\dfrac{\partial P}{\partial y}\neq\dfrac{\partial Q}{\partial x}$ on $D$ → not conservative on $D$.
- $D$ has a hole (e.g. plane minus the origin): $\oint_\gamma\mathbf{v}\cdot d\mathbf{r}$ can be $\neq 0$ on a loop around the hole even if $P_y=Q_x$ away from the singularity.
How to test in 2D (simply connected $D$)
| Step | What to do |
|---|---|
| 1 | Compute $\dfrac{\partial P}{\partial y}$ and $\dfrac{\partial Q}{\partial x}$. If they differ on $D$, stop — no potential on $D$. |
| 2 | If equal, find $\phi$: set $\phi(x,y)=\displaystyle\int P(x,y)\,dx + h(y)$, then $\dfrac{\partial\phi}{\partial y}=Q$ to solve for $h(y)$. |
| 3 | Verify: $\dfrac{\partial\phi}{\partial x}=P$ and $\dfrac{\partial\phi}{\partial y}=Q$. |
Equivalent facts (on simply connected $D$)
- path independence: same endpoints $\Rightarrow$ same line integral;
- $\displaystyle\oint_\gamma \mathbf{v}\cdot d\mathbf{r}=0$ for every closed curve in $D$;
- then $\displaystyle\int_A^B \mathbf{v}\cdot d\mathbf{r}=\phi(B)-\phi(A)$.
Typical mistake Using $\phi(B)-\phi(A)$ without checking $\nabla\phi=\mathbf{v}$, or ignoring holes in the domain.
Example:
Given: $\mathbf{v}(x,y)=(2x,2y)$ on $\mathbb{R}^2$. Goal: evaluate $\displaystyle\int_A^B \mathbf{v}\cdot d\mathbf{r}$ from $A=(0,0)$ to $B=(1,2)$.
Step 1 — Is it conservative? Here $P=2x$, $Q=2y$.
Step 2 — Find $\phi$ (not guess). $\displaystyle\int 2x\,dx=x^2+h(y)$, so $\dfrac{\partial\phi}{\partial y}=h'(y)=2y$, hence $h(y)=y^2$ and $\phi(x,y)=x^2+y^2$ (constant $+C$ does not matter for endpoint difference).
Step 3 — Verify (this is what makes it legal):
Step 4 — Use potential formula:
Check: any path from $(0,0)$ to $(1,2)$ in $\mathbb{R}^2$ gives $5$ because $\mathbf{v}$ is a gradient on the whole plane.
Deeper synthesis (equivalence list, path-integral route, holes): Chapter 12.
10.2 Integration Rules #
Computational rules for antiderivatives. FTC (§10.1.2) turns these into definite integrals once you have an antiderivative.
10.2.1 Integration by Parts #
Idea
Reverse the product rule: $(uv)'=u'v+uv'$, so $uv'$ is integrated by trading a derivative from $u$ to $v$. Choose $u$ so $du$ is simpler (polynomial → lower degree, $\ln x$ → $1/x$, inverse trig → simpler form).
Definite form:
When to use Product integrands where one factor simplifies by differentiation (polynomial, $\ln x$, $\arctan x$, etc.).
When not to use Both factors stay equally messy under differentiation/integration — try substitution or rewrite the integrand first.
Typical mistake Dropped boundary term $\big[uv\big]_a^b$ in definite integrals.
Example:
Goal: $\displaystyle\int_0^1 x\ln(1+x)\,dx$.
Choice: $u=\ln(1+x)$, $dv=x\,dx$ → $du=\frac{dx}{1+x}$, $v=\frac{x^2}{2}$.
Algebra: $\frac{x^2}{1+x}=x-1+\frac{1}{1+x}$ gives $\displaystyle\frac14$.
Check: boundary at $x=0$ vanishes; result $=\frac14>0$ is plausible on $[0,1]$.
10.2.2 Substitution Rule (u-substitution) #
Idea
The integrand already looks like (derivative of inside) $\times$ (function of inside) — the chain rule in reverse. Spot $u=h(x)$ and rewrite $h'(x)\,dx$ as $du$.
If $u=h(t)$ , $du=h'(t)\,dt$ , then
Definite form:
When to use When the integrand has an inner function and (up to constants) its derivative is also present:
- chain-rule pattern: $f'(x)\,g(f(x))$
- powers/compositions like $(ax+b)^n$, $e^{h(x)}$, $\sin(h(x))$, $\cos(h(x))$
When not to use No inner function whose derivative (up to a constant) appears in the integrand — try parts, partial fractions, or a trig rewrite.
Typical mistake Changing variable but not changing bounds on definite integrals.
Example:
Goal: $\displaystyle\int_0^1 \frac{2x}{(1+x^2)^2}\,dx$.
Substitution: $u=1+x^2$, $du=2x\,dx$, bounds $u:1\to2$.
Check: $u$-bounds match $x=0,1$; integrand was exact chain-rule form $g(u)\,du$.
10.2.3 Important Linear Change of Variable #
Idea
A shifted/scaled inner function $at+b$ only stretches the $x$-axis by $1/a$ — that factor appears explicitly in the integral.
For $u=at+b$ :
Definite form:
Typical mistake Forgetting the factor $1/a$ .
Example:
Compute
Set $u=3x-1$ , $du=3dx$ , $dx=\frac{du}{3}$ , bounds $-1\to5$ :
10.3 Frequent Integrals and Standard Methods #
Method map (read this first)
| If the integrand looks like… | Try first… | See |
|---|---|---|
| $f'(x)\,g(f(x))$ or $\frac{f'(x)}{f(x)}$ | substitution | §10.2.2 |
| polynomial $\times$ $e^{ax}$, $\sin$, $\cos$, $\ln$ | integration by parts (repeat) | §10.2.1, §10.3.3 |
| $e^{ax}\sin(bx)$ or $e^{ax}\cos(bx)$ | parts twice, solve for $I$ | §10.3.4 |
| rational function $\frac{P}{Q}$ | divide if $\deg P\ge\deg Q$, then partial fractions | §10.3.7 |
| $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, $\sqrt{x^2-a^2}$ | trig substitution | standard tables |
| rational in $\sin x,\cos x$ | half-angle $t=\tan(x/2)$ | §10.3.8 |
| $\int_a^b f\,ds$ or $\mathbf{F}\cdot d\mathbf{r}$ on a curve | parametrize $\gamma$ | §10.1.8 |
If two methods seem possible, pick the one that reduces degree or algebraic complexity in one clear step.
10.3.1 Elementary Integrals #
Typical mistake Missing absolute value in logarithms.
Example:
Compute
Antiderivative:
Evaluate:
10.3.2 Integrals Linked to Inverse Functions #
Typical mistake Confusing $1/(1-x^2)$ with $1/\sqrt{1-x^2}$ .
Example:
Compute
Since $\int \frac{1}{1-x^2}dx=\operatorname{artanh}(x)$ :
10.3.3 Power Times Exponential or Trigonometric Function #
Idea
Differentiate the polynomial repeatedly until it becomes a constant; each step of parts lowers the power of $x$.
Examples:
- $\int x^n e^{ax}\,dx$ - $\int x^n\sin(ax)\,dx$ , $\int x^n\cos(ax)\,dx$
Method Repeated integration by parts:
- differentiate polynomial part,
- integrate exp/trig part,
- repeat until polynomial vanishes.
Typical mistake Stopping before recognizing the recursion pattern.
Example:
Compute
By parts twice:
10.3.4 Exponential-Trigonometric Products #
Idea
One integration by parts is not enough — the exponential and trig factors “swap roles” but the integral never fully disappears. After two parts steps, the same integral $I$ reappears on the right; then you solve for $I$ like a linear equation.
Standard integrals (template)
Same workflow for both; only the trig factor in the first parts choice changes.
Recommended first choice (cosine case)
| $u$ | $dv$ | |
|---|---|---|
| 1st parts on $I$ | $\cos(bx)$ | $e^{ax}\,dx$ |
| 2nd parts on $\int e^{ax}\sin(bx)\,dx$ | $\sin(bx)$ | $e^{ax}\,dx$ |
Then (schematically, before solving for $I$):
Substitute the second line into the first — the $I$-terms collect on one side.
Algebra pattern (what you are solving)
Here $A(x),B(x)$ are the non-integral terms produced by the two parts steps (products of $e^{ax}$ and trig functions). Do not stop when you still see $\int e^{ax}\cos(bx)\,dx$ or $\int e^{ax}\sin(bx)\,dx$ on the right — finish the second parts, substitute, then solve.
Closed form (check or exam shortcut when $a^2+b^2\neq 0$)
For $a=b=1$: $\displaystyle\int e^x\cos x\,dx=\frac{e^x}{2}(\cos x+\sin x)+C$.
When to use Integrand is $e^{ax}\times$($sin$ or $cos$) — no substitution reduces both factors at once.
When not to use Only one factor is exponential and the other is a polynomial → use repeated parts (§10.3.3). Pure trig products → use identities (§10.3.5).
Typical mistake
- Stopping after one parts step while $\int e^{ax}\sin(bx)\,dx$ (or the cos twin) is still on the right.
- Sign error on $du$ when $u=\cos(bx)$: $du=-b\sin(bx)\,dx$ introduces a plus in front of the second integral.
- Forgetting to move $I$ to the left: $I = \cdots - I$ means $2I=\cdots$, not $I=\cdots$.
Example:
Goal: $I=\displaystyle\int e^x\cos x\,dx$ (here $a=b=1$).
1st parts: $u=\cos x$, $dv=e^x\,dx$ → $du=-\sin x\,dx$, $v=e^x$.
2nd parts on $J=\int e^x\sin x\,dx$: $u=\sin x$, $dv=e^x\,dx$.
Substitute and solve:
Check: matches the formula with $a=b=1$; derivative gives $e^x\cos x$.
10.3.5 Powers of Trigonometric Functions #
Use identities:
General strategy:
- reduce powers,
- rewrite product as sum if needed,
- integrate term-by-term.
Typical mistake Expanding blindly instead of using identities first.
Example:
Compute
Use identity
Integrate over full period, oscillatory terms vanish:
10.3.6 Orthogonal Relations (Fourier Basics) #
Over $[0,2\pi]$ :
Meaning Different frequencies are orthogonal on a full period.
Example:
For $m\neq n$ , compute
Product-to-sum:
Both cosine terms integrate to $0$ over $0$ to $2\pi$ , so the integral is $0$ . This is the orthogonality used in Fourier coefficients.
10.3.7 Rational Integrals (Worked-Example Style) #
Idea
Write $\frac{P}{Q}$ so $\deg P<\deg Q$ (polynomial division if needed). Split the denominator into factors; each piece becomes a log or an arctan after completing the square.
For integrals like
typical workflow:
- Split numerator to include derivative of denominator ( $2x+1$ ).
- Substitution for logarithmic part.
- Complete square for remaining part:
- Use arctangent form.
Typical mistake Trying one substitution only; many problems need decomposition first.
Example:
Goal: $\displaystyle\int \frac{2x+3}{x^2+2x+5}\,dx$.
Step 0 — Name denominator and its derivative
The trick: rewrite the numerator so one piece is exactly $D'(x)$ (gives a log), the rest is a constant over completed square (gives arctan).
Step 1 — Split the numerator
So
Step 2 — Log part ($\dfrac{D'}{D}$)
With $u=D(x)=x^2+2x+5$, we have $du=(2x+2)\,dx=D'(x)\,dx$:
($D(x)>0$ for all real $x$, so $|u|=u$.)
Step 3 — Complete the square on the remaining piece
So the second integral is
Step 4 — Arctan part (shift $t=x+1$)
Let $t=x+1$, $dt=dx$. Use $\displaystyle\int \frac{1}{t^2+a^2}\,dt=\frac{1}{a}\arctan\!\left(\frac{t}{a}\right)$ with $a=2$:
Step 5 — Add both parts
Check: $\dfrac{d}{dx}\ln(x^2+2x+5)=\dfrac{2x+2}{x^2+2x+5}$ matches the $D'/D$ piece; derivative of $\dfrac{1}{2}\arctan\!\left(\dfrac{x+1}{2}\right)$ gives $\dfrac{1}{(x+1)^2+4}=\dfrac{1}{D(x)}$ on the constant-numerator piece.
10.3.8 Tangent Half-Angle Substitution #
Idea
Any expression built from $\sin x$, $\cos x$, and $dx$ that is rational (fractions of polynomials in sin/cos) can be turned into an ordinary rational function in a new variable $t$. The standard choice is $t=\tan(x/2)$ (Weierstrass substitution): $\sin x$, $\cos x$, and $dx$ all become rational functions of $t$.
Substitution formulas (set $t=\tan(x/2)$)
Formulas: §15.6.
Useful combinations (after common denominator $1+t^2$)
Workflow
- Replace $\sin x$, $\cos x$, $dx$ by the formulas above.
- Simplify — the factor $1+t^2$ in $dx$ often cancels part of the denominator.
- Integrate the resulting rational function in $t$ (partial fractions / complete the square / log–arctan).
- Definite integrals: change $x$-bounds to $t$-bounds via $t=\tan(x/2)$ before integrating.
- Indefinite integrals: at the end, back-substitute $t=\tan(x/2)$ (or use a known $t$-formula).
When to use $\displaystyle\int R(\sin x,\cos x)\,dx$ where $R$ is a rational function (polynomials in $\sin x,\cos x$ divided by each other), and simpler methods (power identities, $\int f(\sin x)\cos x\,dx$, etc.) do not apply.
When not to use
- $\int \sin^n x\cos^k x\,dx$ with an odd power → use §10.3.5 instead.
- $\int f(\sin x)\cos x\,dx$ or $\int f(\cos x)\sin x\,dx$ → simple $u$-sub (§10.2.2).
- Integrands with $\tan x$, $\sec x$ only — sometimes a different $u=\tan x$ or $\sec x$ sub is faster.
Typical mistake
- Forgetting $dx=\dfrac{2}{1+t^2}\,dt$ (missing the factor $2/(1+t^2)$).
- Leaving $x$-bounds on a definite integral after switching to $t$.
- Dropping the domain restriction: $t=\tan(x/2)$ is smooth on $(-\pi,\pi)$; pick the branch that contains your interval.
Reference integral (complete the square on $1+2t-t^2=2-(t-1)^2$)
Example:
Goal: $\displaystyle\int_0^{\pi/2}\frac{dx}{\sin x+\cos x}$.
Step 1 — Substitute $t=\tan(x/2)$:
Step 2 — Rewrite the denominator
Step 3 — Simplify the integral ($1+t^2$ cancels)
Step 4 — $x$-bounds $\to$ $t$-bounds
$x=0 \Rightarrow t=\tan 0=0$.
$x=\dfrac{\pi}{2} \Rightarrow t=\tan\dfrac{\pi}{4}=1$.
Step 5 — Integrate in $t$ (square completion)
With $u=t-1$, use $\displaystyle\int\frac{du}{a^2-u^2}=\frac{1}{2a}\ln\left|\frac{a+u}{a-u}\right|$ and $a=\sqrt2$:
At $t=1$: $t-1=0$ gives $\ln 1=0$.
At $t=0$: $t-1=-1$ gives $\dfrac{1}{\sqrt2}\ln\dfrac{\sqrt2-1}{\sqrt2+1}
= -\dfrac{1}{\sqrt2}\ln\dfrac{\sqrt2+1}{\sqrt2-1}$.
Upper minus lower:
Check: integrand $>0$ on $(0,\pi/2)$, so the integral should be positive; $\dfrac{\sqrt2+1}{\sqrt2-1}>1$ and the log is positive.
10.3.9 Solids of Revolution #
Idea
Start with a flat region in the plane (under $y=f(x)$ between $x=a$ and $x=b$). Spin that region around an axis — the solid that appears is the body of revolution. Volume = “sum” of infinitely thin slices; the only hard part is picking the slice shape so each piece matches the axis of rotation.
Step 0 — Picture the 2D region
You always begin with the same generating region: $\{(x,y): a\le x\le b,\ 0\le y\le f(x)\}$ (or between two curves). The plots below use $f(x)=x^2$ on $[0,1]$ — a curved region (not a triangle), so disk radii and shell heights look different in 2D.
Disk method (slice ⊥ to the x-axis) — rotate around the x-axis
- Slice: a thin vertical strip at position $x$ becomes a disk (pancake) perpendicular to the $x$-axis.
- Radius: distance from the $x$-axis to the curve = $f(x)$ (because the axis is the $x$-axis).
- Thickness: $dx$.
- Volume element: $dV = \pi [\text{radius}]^2\,dx = \pi [f(x)]^2\,dx$.
Disk + 3D (rotation about the $x$-axis): Left — vertical segments are disk radii $r=f(x)$ (slice ⊥ to $x$); notice they shrink toward $x=0$ because $x^2$ is flat there. Right — drag to rotate ($V=\pi/5$ here); red curve = profile in $z=0$.
Shell method (slice ‖ to the y-axis) — rotate around the y-axis
- Slice: a thin vertical strip at position $x$ becomes a cylindrical shell (label on a soup can) with axis parallel to the $y$-axis.
- Radius: horizontal distance from the $y$-axis = $x$.
- Height: $f(x)$ (top of the region at that $x$).
- Thickness: $dx$.
- Volume element: $dV = 2\pi(\text{radius})(\text{height})\,dx = 2\pi x\,f(x)\,dx$.
Shell + 3D (rotation about the $y$-axis): Left — red = shell radius $x$ (horizontal); green = height $f(x)=x^2$ (vertical). At $x=0.75$ the radius is longer than the height — an L-shape, not a single segment like the disk plot. Right — same profile spun about $y$ ($V=\pi/2$).
Which method for which axis?
| Rotate around… | Slice perpendicular to… | Formula | Radius | Height (if shell) |
|---|---|---|---|---|
| x-axis | $x$ (disk) | $\pi\int_a^b [f(x)]^2\,dx$ | $f(x)$ | — |
| y-axis | $y$ (shell, usual setup) | $2\pi\int_a^b x\,f(x)\,dx$ | $x$ | $f(x)$ |
Same region, different axis → different solid and different volume ($\pi/5$ vs $\pi/2$ for $f(x)=x^2$ on $[0,1]$ below).
Washer method (hole in the middle, axis = $x$):
where $R(x)$ = outer radius, $r(x)$ = inner radius (both measured perpendicular to the $x$-axis). Use when the slice at $x$ is an annulus (disk with a hole), not a solid disk.
Example region: between $r(x)=x^2$ and $R(x)=1$ on $[0,1]$ — rotate about the $x$-axis. Left: outer radius $R(x)$ (red) and inner $r(x)$ (teal) at sample $x$. Right: hollow solid (outer and inner surfaces).
Washer volume
Region between $r(x)=x^2$ and $R(x)=1$ on $[0,1]$, axis = $x$:
When not to use
- Disk formula with rotation about the y-axis without rewriting $x$ as a function of $y$ — radius is not $f(x)$.
- Shell formula with rotation about the x-axis using $2\pi x f(x)$ — wrong radius (use disks, or shells with thickness $dy$).
Typical mistake Using $f(x)$ as radius when the axis of rotation is the y-axis (radius is $x$, not $y$).
Example:
Region: under $f(x)=x^2$ on $[0,1]$ (see first plot).
A) Rotate around the $x$-axis (disk method)
Slice at $x$: disk radius $r=f(x)=x^2$, so $dV=\pi (x^2)^2\,dx=\pi x^4\,dx$.
Solid: a paraboloid-type body (profile flattens near $x=0$, widest at $x=1$).
B) Rotate the same region around the $y$-axis (shell method)
Slice at $x$: shell radius $x$, height $f(x)=x^2$, so $dV=2\pi x\cdot x^2\,dx=2\pi x^3\,dx$.
Solid: a different body (same parabolic edge, spun about the vertical axis) — larger volume than (A).
Check: $V_y=\tfrac{5}{2}V_x$ here; compare the 2D slice diagrams (disk = vertical radii only; shell = horizontal radius + vertical height, usually unequal when $f$ is not linear).
10.4 Integration Exercises (End of Chapter 10) #
Decide first which method fits best — use the method map in §10.3 and the checklist below. Stuck? Riemann limits → §10.1.3; FTC → §10.1.2; bounds after $u$-sub → §10.2.2.
A) Use u-substitution #
- $\displaystyle \int_0^1 \frac{3x^2}{1+x^3}\,dx$
- $\displaystyle \int x\,e^{x^2}\,dx$
- $\displaystyle \int \frac{\cos(\ln x)}{x}\,dx$
- $\displaystyle \int_0^{\pi/4}\sec^2(2x)\,dx$
- $\displaystyle \int \frac{dx}{\sqrt{1-4x^2}}$
B) Use integration by parts #
- $\displaystyle \int x e^x\,dx$
- $\displaystyle \int \ln x\,dx$
- $\displaystyle \int x\cos x\,dx$
- $\displaystyle \int_0^1 x^2\ln(1+x)\,dx$
- $\displaystyle \int e^x\sin x\,dx$
C) Mixed practice (choose method) #
- $\displaystyle \int \frac{x^2+1}{x+1}\,dx$
- $\displaystyle \int_0^2 \frac{x}{(1+x^2)^2}\,dx$
- $\displaystyle \int \frac{1}{x^2-1}\,dx$
- $\displaystyle \int \frac{dx}{\sqrt{x^2+4}}$
- $\displaystyle \int_0^{\pi/2}\frac{dx}{1+\sin x}$
Quick decision checklist #
- Try u-substitution when you see "inside function + derivative of inside" (§10.2.2).
- Try integration by parts for products where differentiating one factor simplifies it (§10.2.1, §10.3.3–10.3.4).
- Try partial fractions for rational functions after polynomial division if needed (§10.3.7).
- Try trig substitution for $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, $\sqrt{x^2-a^2}$.
- For limits of sums, run the Riemann checklist: $\sum f(x_k)\Delta x$ and $b-a=n\Delta x$ (§10.1.3).
- For improper integrals, split at singularities/infinity and take limits separately (§10.1.5).
Answers (Final Results) #
A) Use u-substitution
- $\displaystyle \int_0^1 \frac{3x^2}{1+x^3}\,dx=\ln 2$
- $\displaystyle \int x\,e^{x^2}\,dx=\frac12 e^{x^2}+C$
- $\displaystyle \int \frac{\cos(\ln x)}{x}\,dx=\sin(\ln x)+C$
- $\displaystyle \int_0^{\pi/4}\sec^2(2x)\,dx$ diverges (improper integral, $=+\infty$)
- $\displaystyle \int \frac{dx}{\sqrt{1-4x^2}}=\frac12\arcsin(2x)+C$
B) Use integration by parts
- $\displaystyle \int x e^x\,dx=e^x(x-1)+C$
- $\displaystyle \int \ln x\,dx=x\ln x-x+C$
- $\displaystyle \int x\cos x\,dx=x\sin x+\cos x+C$
- $\displaystyle \int_0^1 x^2\ln(1+x)\,dx=\frac{2}{3}\ln 2-\frac{5}{18}$
- $\displaystyle \int e^x\sin x\,dx=\frac12 e^x(\sin x-\cos x)+C$
C) Mixed practice
- $\displaystyle \int \frac{x^2+1}{x+1}\,dx=\frac{x^2}{2}-x+2\ln|x+1|+C$
- $\displaystyle \int_0^2 \frac{x}{(1+x^2)^2}\,dx=\frac{2}{5}$
- $\displaystyle \int \frac{1}{x^2-1}\,dx=\frac12\ln\left|\frac{x-1}{x+1}\right|+C$
- $\displaystyle \int \frac{dx}{\sqrt{x^2+4}}=\operatorname{arsinh}\!\left(\frac{x}{2}\right)+C$
- $\displaystyle \int_0^{\pi/2}\frac{dx}{1+\sin x}=1$
Worked Solutions (Step by Step) #
A1 $\displaystyle \int_0^1 \frac{3x^2}{1+x^3}\,dx$
A2 $\displaystyle \int x\,e^{x^2}\,dx$
A3 $\displaystyle \int \frac{\cos(\ln x)}{x}\,dx$
A4 $\displaystyle \int_0^{\pi/4}\sec^2(2x)\,dx$
For the definite integral:
As $b\to\pi/4^-$, $2b\to\pi/2^-$ and $\tan(2b)\to+\infty$, so the integral diverges.
A5 $\displaystyle \int \frac{dx}{\sqrt{1-4x^2}}$
B1 $\displaystyle \int x e^x\,dx$
B2 $\displaystyle \int \ln x\,dx$
B3 $\displaystyle \int x\cos x\,dx$
B4 $\displaystyle \int_0^1 x^2\ln(1+x)\,dx$
Polynomial division:
So
Therefore
B5 $\displaystyle \int e^x\sin x\,dx$ Let
First integration by parts:
Set
Again by parts:
Substitute into $I$:
C1 $\displaystyle \int \frac{x^2+1}{x+1}\,dx$ Polynomial division:
Hence
C2 $\displaystyle \int_0^2 \frac{x}{(1+x^2)^2}\,dx$
C3 $\displaystyle \int \frac{1}{x^2-1}\,dx$ Partial fractions:
So
C4 $\displaystyle \int \frac{dx}{\sqrt{x^2+4}}$ Use $x=2\tan\theta$:
Then
Back-substitute:
so
C5 $\displaystyle \int_0^{\pi/2}\frac{dx}{1+\sin x}$ Rationalize:
Therefore
Take the limit at $\pi/2$:
and at $0$:
Hence
Chapter 11 — Integration Multidimensional $\mathbb{R}^n$ #
Multiple integrals, change of variables (Jacobian), Green's theorem, surface and volume integrals, center of mass, and the divergence theorem (Gauss) — with exam-style practice at the end.
Story of the chapter: in $\mathbb{R}$ we integrated on intervals (Chapter 10). Here the domain is a set $\Omega\subset\mathbb{R}^n$. We approximate $\Omega$ by boxes (Jordan measure), evaluate integrals by Fubini (iterated 1D integrals) on rectangles and normal regions, then change coordinates when the shape is a disk, ball, or cylinder (Jacobian). Line integrals from §10.1.8–10.1.9 connect to Green and Poincaré; flux through a surface leads to Gauss.
Prerequisites from Chapter 10: definite integrals, FTC, line integrals $\int_\gamma \mathbf{v}\cdot d\mathbf{r}$, conservative fields ($\nabla\phi=\mathbf{v}$, $P_y=Q_x$).
Also uses: partial derivatives and Jacobian (Ch 8), parameter surfaces as in Ch 9 / Ch 13.
Theorem ladder (how the pieces fit)
| Step | Tool | Converts |
|---|---|---|
| 1 | Fubini + normal regions (§11.1–11.2) | Box-like domains → iterated 1D integrals |
| 2 | Jacobian / polar / spherical (§11.3) | Curved domains → simpler coordinates |
| 3 | Green (§11.4) | Double integral over $\Omega$ ↔ line integral on $\partial\Omega$ |
| 4 | Surface / flux integrals (§11.5) | Fields on curved surfaces in $\mathbb{R}^3$ |
| 5 | Gauss (§11.8) | Flux through closed $\partial V$ ↔ volume integral of $\mathrm{div}$ |
Green is the 2D case of Stokes (Ch 13 §13.0); Gauss is the 3D divergence form of the same “interior cancels, boundary remains” idea.
11.1 Definitions and Foundations #
11.1.1 Boxes, elementary figures, and Jordan measure #
Idea
In $\mathbb{R}^n$, a box is a product of intervals $Q=[a_1,b_1]\times\cdots\times[a_n,b_n]$. An elementary figure is a finite union of boxes (disjoint up to shared faces). A bounded set $\Omega$ is Jordan measurable if its boundary has zero $(n-1)$-dimensional "area" — then we can trap $\Omega$ between elementary figures $E\subset\Omega\subset G$ with $\mu(G\setminus E)$ arbitrarily small. Full topology link: §14.7.
Read $\mu(\partial\Omega)=0$: the boundary is so thin that it contributes no area (in $\mathbb{R}^2$) or volume (in $\mathbb{R}^3$) in the limit of box approximations — a circle boundary is fine; a boundary filled with a 2D dust of rationals is not.
Volume of a box
Riemann integral on a box (same idea as $\mathbb{R}$): partition $Q$ into sub-boxes, use step functions $s\le f\le h$, take infimum of upper sums and supremum of lower sums; if they agree, that value is $\int_Q f\,d\mu$.
Jordan measurable: bounded $\Omega\subset\mathbb{R}^n$ is Jordan measurable iff
equivalently: for every $\varepsilon>0$ there exist elementary figures $E,G$ with $E\subset\Omega\subset G$ and $\mu(G\setminus E)<\varepsilon$.
Typical mistake Confusing "bounded" with "Jordan measurable" — a set can be bounded but have a thick fractal boundary (not Jordan measurable in the classical sense).
Example:
The unit disk $\{(x,y):x^2+y^2\le 1\}$ is Jordan measurable (circle boundary has measure $0$ in $\mathbb{R}^2$). A filled square $[0,1]^2$ is an elementary figure, hence Jordan measurable.
11.1.2 Content and notation #
On a Jordan measurable set $\Omega$ and integrable $f$:
All denote the same object. In $\mathbb{R}^2$ we often write $dA$ or $dx\,dy$; in $\mathbb{R}^3$, $dV$.
11.1.3 Fubini's theorem (iterated integrals) #
Idea
On a box $Q=[a_1,b_1]\times\cdots\times[a_n,b_n]$, if $f$ is continuous on $Q$ (or at least integrable in the Riemann sense used in your course), you may compute the integral as an $n$-fold iterated integral — the order of integration does not matter.
Theorem (Fubini on a box) Let $Q=[a_1,b_1]\times\cdots\times[a_n,b_n]$ and $f:Q\to\mathbb{R}$ continuous. Then
and any permutation of the integration order gives the same value.
When not to use
- Domain is not a box — first describe $\Omega$ as a normal region (§11.2) or change variables (§11.3).
- $f$ is not continuous on the closure — singularities need splitting (as in improper integrals, §10.1.5).
Typical mistake Wrong bounds when setting up the inner integral (the outer variable must describe the full slice of $\Omega$).
Example:
Compute $\displaystyle\int_Q (x+y)\,d\mu$ on $Q=[0,1]\times[0,2]$.
Method: Fubini — integrate in $y$ first:
Read the numbers: the flat rectangle $Q$ has geometric area $1\times 2=2$. The value $3$ is the volume under the graph $z=f(x,y)=x+y$ over $Q$ (not the area of $Q$).
Fubini picture: for fixed $x$, integrate $(x+y)$ in $y$ from $0$ to $2$ — that is the area of a vertical slice (height varies linearly in $y$). You get slice area $2x+2$; integrating that in $x$ from $0$ to $1$ stacks the slices to total volume $3$.
Check: integrate in $x$ first — same answer $3$.
Finer partitions of $Q$ make the boxed approximation fill the volume under the plane; the limit of those sums is $3$.
11.1.4 Domains of class $C^1$, $C^{1}_{\mathrm{pw}}$, $C^k$ #
Idea
A $C^1$ domain is a bounded open set $\Omega$ whose boundary $\partial\Omega$ is a smooth surface (no sharp corners, no cusps). Locally, $\Omega$ is always the same shape: “all points below one smooth graph,” written in suitable coordinates.
What you should picture (2D first)
In $\mathbb{R}^2$, near a boundary point $p$ you want a small window where $\Omega$ looks like
with $\psi$ a smooth function (no corner on $\partial\Omega$ inside that window). The formal definition in $\mathbb{R}^n$ is exactly this idea, with the last coordinate playing the role of $y$ and the first $n-1$ coordinates bundled as $x_0$.
Definition (graph over a quader — unpacked)
$\Omega\subset\mathbb{R}^n$ is a $C^1$ domain if for every boundary point $p\in\partial\Omega$ the following local description exists:
| Piece | Meaning |
|---|---|
| Split $\mathbb{R}^n=\mathbb{R}^{n-1}\times\mathbb{R}$ | Write a point as $(x_0,x_n)$: $x_0=(x_1,\ldots,x_{n-1})$, $x_n$ is the last coordinate (you may rotate coordinates so the boundary is not vertical in the last direction). |
| Box $Q_0\subset\mathbb{R}^{n-1}$ | A small rectangle in the $(n-1)$ variables (the “Quader” / brick in the base). |
| Interval $]c,d[$ | An open range for $x_n$ large enough to contain $p$. |
| Neighborhood $W=Q_0\times]c,d[$ | A small open box around $p$. |
| $\psi\in C^1(Q_0)$ | A once continuously differentiable function describing the boundary as a graph. |
Inside $W$, $\Omega$ is exactly the set of points strictly below that graph:
How to read the inequality $c<x_n<\psi(x_0)$
- For each fixed $x_0$ in the base box $Q_0$, the last coordinate $x_n$ may vary in an interval between a floor $c$ and a ceiling $\psi(x_0)$.
- The upper boundary $\partial\Omega$ is the graph $x_n=\psi(x_0)$ (smooth).
- The lower part in this window is flat at $x_n=c$ (that face need not be part of $\partial\Omega$ globally — it is just the bottom of the local box $W$).
At a different boundary point you may need a different coordinate split (e.g. “below a graph in $x$” instead of “below a graph in $y$”). That is allowed: the definition only asks that some smooth graph description works near each $p$.
Example in $\mathbb{R}^2$
Let $\Omega=\{(x,y): 0<x<1,\ x^2<y<1\}$ (region between the parabola $y=x^2$ and the line $y=1$).
Near a point on the top edge $y=1$, use coordinates $(x_0,y)=(x,y)$, $Q_0=(0,1)$, $\psi(x)=1$, and choose $c$ small enough so $c<y<1=\psi(x)$. Then $\Omega\cap W$ is “$x$ in an interval, $y$ below the graph $\psi(x)=1$.”
Near a point on the curved edge $y=x^2$, the same coordinates work with $\psi(x)=x^2$ and $c$ below the parabola: $\Omega\cap W$ is “$y$ below the smooth graph $x^2$.”
Near a corner $(1,1)$ where $y=1$ and $y=x^2$ meet: a single graph $\psi$ cannot describe both edges in one $W$ — you split into two neighborhoods (one using “below $y=1$”, one using “above $y=x^2$” in swapped coordinates). That is why the full square $[0,1]^2$ is only $C^1_{\mathrm{pw}}$ (piecewise smooth), not $C^1$.
What is NOT a $C^1$ domain
- A set whose boundary has a reentrant corner (like an inward cusp): near the cusp, no smooth graph $x_n=\psi(x_0)$ describes only one side of $\Omega$.
- “Swiss cheese” with fractal boundary (Jordan measurable can still hold, but not $C^1$).
$C^{1}_{\mathrm{pw}}$ (piecewise $C^1$)
Same local picture, but $\psi$ may be glued from finitely many smooth pieces. A square or rectangle is $C^1_{\mathrm{pw}}$: each edge is a smooth graph in some coordinates; the corners are handled by changing coordinates from edge to edge.
$C^k$
The boundary is described by $\psi\in C^k$ (more derivatives). Stronger regularity for sharper theorems; exams often stop at $C^1$ or $C^1_{\mathrm{pw}}$.
Why it matters
Green's theorem, Gauss's theorem, and the change-of-variables formula are proved for domains whose boundary is well behaved (oriented, piecewise smooth). $C^1$ / $C^1_{\mathrm{pw}}$ is the usual hypothesis so surface integrals $\int_{\partial\Omega}$ and volume integrals $\int_\Omega$ can be related without pathological boundaries.
Typical mistake
- Confusing $C^1$ domain with Jordan measurable: a disk is both; a region with a cusp can still be Jordan measurable but is not $C^1$.
- Applying Green/Gauss on a domain with a hole (annulus) without the inner boundary: $\partial\Omega$ has two closed curves, both oriented with $\Omega$ on the left.
11.2 Integration over Normal Regions #
Idea
A normal region (Normalbereich) is a set where one variable runs in an interval and the others are trapped between graphs of functions — Fubini becomes a concrete 2D (or 3D) iterated integral with explicit bounds.
Type I in $\mathbb{R}^2$ ($x$ outer) #
Then
Type II in $\mathbb{R}^2$ ($y$ outer) #
Then
Method choice
- If vertical slices $\{x=\text{const}\}\cap\Omega$ are always intervals $[f(x),g(x)]$ → Type I.
- If horizontal slices are nicer → Type II.
- Sometimes split $\Omega$ into pieces of different types.
Typical mistake Using $f(x)\le y\le g(x)$ when the region is not a single $x$-strip (e.g. circle needs polar coordinates or two Type-I pieces).
Region for many examples below: between $y=x^2$ and $y=1$ on $x\in[0,1]$.
Picture: walking in $x$ (Type I) #
For Type I you fix $x$ and let $y$ run from the lower graph $f(x)$ to the upper graph $g(x)$. Then you move $x$ from $a$ to $b$ and add up slice areas.
- At $x=0.5$: $f(0.5)=0.25$, $g(0.5)=1$ → strip height $L(x)=g(x)-f(x)=0.75$.
- A thin slab of width $\Delta x$ has area $\approx L(x)\,\Delta x$; shrinking $\Delta x$ makes that rectangle match the true slice (Riemann step toward Fubini).
Type II on the same region
Here $\Omega$ is also Type II: for fixed $y\in[0,1]$, $x$ runs from $\alpha(y)=\sqrt{y}$ to $\beta(y)=1$. Use Type II when horizontal slices are simpler than vertical ones.
Example (area):
Area of $\Omega=\{0\le x\le 1,\ x^2\le y\le 1\}$.
Check: $1-\frac{1}{3}=\frac{2}{3}$.
11.2.1 Normal regions in $\mathbb{R}^3$ (brief) #
Type: $z$ between two graphs over a 2D region $D$ in the $(x,y)$-plane:
Example setup: unit cylinder $x^2+y^2\le 1$, $0\le z\le h$ — integrate $D$ in polar coordinates (§11.3).
11.3 Change of Variables (Jacobian) #
Idea
A change of variables $\mathbf{x}=\Phi(\mathbf{u})$ distorts area/volume by a factor $|\det D\Phi|$ at each point. You integrate over the new domain $\Phi^{-1}(\Omega)$ in $\mathbf{u}$-coordinates and multiply by that factor.
What you should picture (2D)
At a point $\mathbf{u}_0$, the linear map $D\Phi(\mathbf{u}_0)$ sends a tiny square in the $(u,v)$-plane to a parallelogram in the $(x,y)$-plane. Its area is multiplied by $\lvert\det D\Phi(\mathbf{u}_0)\rvert$. Summing over many small squares in $\tilde\Omega$ and taking the limit gives the factor $|\det D\Phi|$ under the integral — polar $r$ is exactly this: the image of a small $r$–$\varphi$ rectangle is approximately a curved patch of area $r\,dr\,d\varphi$.
Coordinates used in §11.3 (symbol table)
| Name | Parameters | Map to $(x,y)$ or $(x,y,z)$ | Typical ranges |
|---|---|---|---|
| Polar | $(r,\varphi)$ | $x=r\cos\varphi$, $y=r\sin\varphi$ | $r\ge 0$, $\varphi\in[0,2\pi)$ |
| Cylindrical | $(r,\varphi,z)$ | same $x,y$ as polar, $z=z$ | disk in $xy$ × interval in $z$ |
| Spherical (ETH) | $(r,\vartheta,\varphi)$ | $x=r\sin\vartheta\cos\varphi$, $y=r\sin\vartheta\sin\varphi$, $z=r\cos\vartheta$ | $r\ge 0$, $\vartheta\in[0,\pi]$, $\varphi\in[0,2\pi)$ |
Here $\Phi$ sends parameter space (where the integral is easy) to physical space (where the set $\Omega$ lives). The factor $|\det D\Phi|$ (or $r$, $r^2\sin\vartheta$ from the cheat sheet) is the local area/volume scaling.
Theorem (transformation rule, $\mathbb{R}^2$) Let $\Omega\subset\mathbb{R}^2$ be Jordan measurable, $\Phi:U\to\mathbb{R}^2$ a $C^1$ diffeomorphism on a neighborhood of $\bar\Omega$ with $\tilde\Omega=\Phi^{-1}(\Omega)\subset U$. If $f$ is integrable on $\Omega$, then
$3$D: replace $dA$ by $dV$, $|\det D\Phi|$ is the $3\times 3$ Jacobian determinant.
Typical mistake Forgetting $|\det D\Phi|$ or using the wrong sign — always take absolute value for unsigned area/volume integrals.
When not to use
- $\Phi$ is not one-to-one on the domain (e.g. polar at the origin) — restrict to a set where it is bijective, or use symmetry.
- Jacobian vanishes somewhere (fold) — transformation invalid there.
11.3.1 Jacobian cheat sheet #
Use $J=\lvert\det D\Phi\rvert$ and $dA$ or $dV = J\,du\,dv$ (new coordinates on the right).
| Coordinates | $\Phi$ | Geometry (what $\tilde\Omega$ looks like) | $J=\lvert\det D\Phi\rvert$ | $dA$ or $dV$ |
|---|---|---|---|---|
| Polar $(r,\varphi)$ | $x=r\cos\varphi$, $y=r\sin\varphi$ | disk, annulus, sector in $(x,y)$ | $r$ | $r\,dr\,d\varphi$ |
| Cylindrical $(r,\varphi,z)$ | $x=r\cos\varphi$, $y=r\sin\varphi$, $z=z$ | cylinder, prism over a disk | $r$ | $r\,dr\,d\varphi\,dz$ |
| Spherical $(r,\vartheta,\varphi)$ | $x=r\sin\vartheta\cos\varphi$, $y=r\sin\vartheta\sin\varphi$, $z=r\cos\vartheta$ | ball, spherical shell, cone sector | $r^2\sin\vartheta$ | $r^2\sin\vartheta\,dr\,d\vartheta\,d\varphi$ |
Convention note: ETH scripts sometimes use $\vartheta$ = polar angle from $z$-axis and $\varphi$ = azimuth; always match your lecture's symbol table.
Standard volumes (memorize)
| Body | Volume |
|---|---|
| Disk radius $R$ | $\pi R^2$ |
| Ball radius $R$ | $\frac{4}{3}\pi R^3$ |
| Cylinder radius $R$, height $h$ | $\pi R^2 h$ |
| Cone radius $R$, height $h$ | $\frac{1}{3}\pi R^2 h$ |
11.3.2 Polar coordinates ($\mathbb{R}^2$) #
Meaning of $r$ and $\varphi$
- $r\ge 0$ = distance from the origin to $P=(x,y)$.
- $\varphi$ = angle from the positive $x$-axis to the ray $OP$ (counterclockwise).
On that ray, trigonometry in the right triangle gives
Point and triangle
Drag $\varphi$ and $r$ — watch $P$ move. Left: $(x,y)$ with red $r=|OP|$, blue along the $x$-axis ($x=r\cos\varphi$), and green vertical at that $x$ ($y=r\sin\varphi$) — a right triangle on the coordinate axes, not a second pair of axes. Right: the same point in the $(r,\varphi)$ plane.
Why the Jacobian factor is $r$
In the $(r,\varphi)$-plane, a small rectangle $dr\,d\varphi$ maps to a curved patch in $(x,y)$. The arc length in the angular direction is about $r\,d\varphi$ (not $d\varphi$), so area scales by $r$:
Jacobian patch (interactive). Left: the image in $(x,y)$ of a small rectangle in $(r,\varphi)$. Right: that same rectangle in the parameter plane — its area is always $\Delta r\,\Delta\varphi$ (fix $\Delta r$, $\Delta\varphi$ and it does not change).
-
Vary $r_0$, keep $\Delta r$, $\Delta\varphi$ fixed. The rectangle on the right stays the same size; the curved patch on the left grows or shrinks. Closer to the origin ($r_0$ small) the patch has less area; farther out ($r_0$ large) it has more. The caption below the plot reports $\Delta r\,\Delta\varphi$ and an estimate $r_0\,\Delta r\,\Delta\varphi$ for the patch area; their ratio stays $\approx r_0$. That is why $dA\approx r\,dr\,d\varphi$: the same $(dr,d\varphi)$ cell carries more area at larger $r$.
-
Shrink $\Delta r$, $\Delta\varphi$. Both panels show a smaller cell; the ratio in the caption still tracks $r_0$ (the linear approximation improves as the cell gets tiny).
Annulus domain $r_1\le r\le r_2$, $0\le\varphi\le 2\pi$: washer in $(x,y)$ matches the rectangle in $(r,\varphi)$. Use the r and φ sliders to see the same point in both planes — inner/outer circles correspond to $r=r_1$ and $r=r_2$; the full $\varphi$ edge is one full turn.
Example (disk):
$\displaystyle\int_D (x^2+y^2)\,dA$ on $D=\{x^2+y^2\le 1\}$.
Example (washer region from §10.3.9):
Between $r=x^2$ and $R=1$ about $x$-axis — in the plane, the generating region $\{0\le x\le 1,\ x^2\le y\le 1\}$ has area $\frac{2}{3}$ (§11.2). Polar is useful when $\Omega$ is rotationally symmetric.
11.3.3 Cylindrical and spherical ($\mathbb{R}^3$) #
Compact $dA$, $dV$ factors: §15.12.
Cylindrical $(r,\varphi,z)$
Same horizontal map as polar; the $z$-coordinate is unchanged. A vertical column above a point in the disk is a line segment in $z$.
Cylinder $x^2+y^2\le R^2$, $0\le z\le h$:
Spherical $(r,\vartheta,\varphi)$
- $r\ge 0$ = distance from origin.
- $\vartheta\in[0,\pi]$ = polar angle from the $+z$-axis (colatitude; $\vartheta=0$ is north pole, $\vartheta=\pi$ is south pole).
- $\varphi\in[0,2\pi)$ = azimuth in the $xy$-plane (same role as polar $\varphi$).
Project $P$ onto the $xy$-plane: horizontal radius is $r\sin\vartheta$, then polar formulas give
The volume element is $r^2\sin\vartheta\,dr\,d\vartheta\,d\varphi$ (factor $r^2\sin\vartheta$ from stretching in $r$, $\vartheta$, and $\varphi$).
Jacobian patch (interactive). Left (3D): the curved shell in $(x,y,z)$ for the parameter brick $[r_0,r_0+\Delta r]\times[\vartheta_0,\vartheta_0+\Delta\vartheta]\times[\varphi_0,\varphi_0+\Delta\varphi]$ — inner cap at $r_0$, outer at $r_0+\Delta r$, four side faces, plus a faint reference sphere at $r=1.2$; axes span the ball but the default view is zoomed on the patch (drag to rotate, scroll out to see the full sphere). Right (3D): the same cell as an ordinary axis-aligned box in the full $(r,\vartheta,\varphi)$ domain — the wireframe brick moves when you change $r_0$, $\vartheta_0$, or $\varphi_0$; $\Delta r$, $\Delta\vartheta$, $\Delta\varphi$ are its edge lengths.
-
Increase $\Delta r$. Left: shell thickens between inner and outer caps. Right: box grows in the $r$ direction. Metrics: $\Delta r\,\Delta\vartheta\,\Delta\varphi$ increases.
-
Vary $\vartheta_0$, keep all $\Delta$ fixed. Right box stays the same shape; left shell changes size ($\sin\vartheta$ factor). Ratio $r_0^2\sin\vartheta_0$ in the caption tracks $\vartheta_0$.
-
Shrink all $\Delta$. Both cells shrink; ratio $\approx r_0^2\sin\vartheta_0$ stays stable (linearization improves).
Ball $x^2+y^2+z^2\le R^2$:
Example (ball volume):
Goal: volume of the ball $B_R=\{(x,y,z):x^2+y^2+z^2\le R^2\}$.
Setup: $f\equiv 1$, so $V=\iiint_{B_R}1\,dV$. In spherical coordinates the ball is $0\le r\le R$, $0\le \vartheta\le \pi$, $0\le \varphi\le 2\pi$, and $dV=r^2\sin\vartheta\,dr\,d\vartheta\,d\varphi$:
Step 1 — $r$-integral (inner): $\sin\vartheta$ is constant in $r$:
Step 2 — $\vartheta$-integral:
Step 3 — $\varphi$-integral:
Combine (constants factorize because the integrand is a product $r^2\cdot\sin\vartheta\cdot 1$):
11.4 Green's Theorem and Poincaré #
Idea
Why only the boundary appears (tiling) — Chop $\Omega$ into many small rectangles. On each cell, Green holds in miniature: the line integral around that cell’s border equals the double integral of $Q_x-P_y$ over the cell. When you add all cells, interior edges are traced twice in opposite directions, so their contributions cancel. Only the outer boundary $\partial\Omega$ remains. That is why a double integral over the whole region can become one line integral on the boundary (and vice versa).
Orientation: walk along $\partial\Omega$ so $\Omega$ lies on your left — then the signs in the cancellation argument match the theorem.
After the statement below, read “Before Green” for what each side integrates, then the worked examples (boundary vs interior on each).
Links: §10.1.8, §10.1.9, §11.8 (box partition), Ch 13 §13.3 (Green as 2D Stokes; full theorem map).
Increase $n$ below: the partition is grid lines only (not a carpet of cell arrows). CCW arrows sit on the outer boundary $\partial\Omega$ only; one highlighted interior edge shows the same segment traced twice in opposite directions (cell A: $+$, cell B: $-$) — those contributions cancel. Only the outer rim survives when you sum all cells.
Green's theorem Let $\Omega\subset\mathbb{R}^2$ be a bounded $C^{1}_{\mathrm{pw}}$ domain, $\partial\Omega$ oriented so $\Omega$ lies on the left as you walk along $\partial\Omega$. Let $P,Q\in C^1(\bar\Omega)$. Then
Before Green: what each side integrates
Vector field — $\mathbf{v}=(P,Q)$ with scalar components $P(x,y)$, $Q(x,y)$.
Right side $\displaystyle\oint_{\partial\Omega}(P\,dx+Q\,dy)=\oint_{\partial\Omega}\mathbf{v}\cdot d\mathbf{r}$ (§10.1.8):
- Integrate on the boundary curve only.
- At each point on $\partial\Omega$: add $P(x,y)\,dx+Q(x,y)\,dy$ (field along the tiny step).
- You integrate $P$ and $Q$ along the path — not curl, not $Q_x-P_y$ on the curve.
Left side $\displaystyle\iint_\Omega(Q_x-P_y)\,dA$:
- Integrate over the filled region $\Omega$.
- Integrand: one scalar $Q_x-P_y=\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}$ from the same $P,Q$.
- Meaning: swirl density inside; total swirl $=\iint_\Omega(Q_x-P_y)\,dA$.
Green: total swirl inside $=$ net circulation on the boundary (equal numbers, different integrands).
Where $P,Q$ come from
- Given in the problem: $\oint(2xy\,dx+x^2\,dy)$ $\Rightarrow$ $P=2xy$, $Q=x^2$.
- Chosen by you (area only): e.g. $P=0$, $Q=x$ so $Q_x-P_y=1$ and $\oint x\,dy=\mathrm{Area}(\Omega)$.
Recipe: (1) Identify $P,Q$ (read or choose). (2) Compute whichever side is easier.
What exam problems look like
| Archetype | Typical wording | $(P,Q)$ | What you usually compute |
|---|---|---|---|
| A — Given field | “Evaluate $\oint_{\partial\Omega}(P\,dx+Q\,dy)$” / circulation of $\mathbf{v}$ | Given | Often $\iint_\Omega(Q_x-P_y)\,dA$ if derivatives are simple |
| B — Area | “Find the area enclosed by …” | You choose e.g. $P=0$, $Q=x$ | $\oint_{\partial\Omega} x\,dy$ on the boundary |
Use Green when: closed boundary and one side is clearly shorter.
Skip Green when: open curve (potential from §10.1.9); conservative on simply connected $\Omega$ ($\oint=0$ without Green).
Real world (one line): circulation of fluid around a loop $=$ total vorticity ($Q_x-P_y$) inside the loop.
Typical mistake
- Treating Green (always a conversion between $\iint_\Omega$ and $\oint_{\partial\Omega}$) like Poincaré (only when $\mathbf{v}=\nabla\phi$ on a simply connected domain).
- Using Poincaré on a region with a hole after checking only $P_y=Q_x$ — circulation around the hole can still be $\neq 0$ (see §11.4.1).
Common form (area): with $P=0$, $Q=x$:
Example (Archetype B — area: choose $(P,Q)$, then integrate on $\partial\Omega$):
Given: $\Omega=\{x^2+y^2\le 1,\ x\ge 0,\ y\ge 0\}$ (quarter unit disk). Asked for: $\mathrm{Area}(\Omega)=\iint_\Omega 1\,dA$.
No field is given — we choose $P=0$, $Q=x$ so $Q_x-P_y=1$. Then Green gives $\mathrm{Area}(\Omega)=\oint_{\partial\Omega} x\,dy=\frac{1}{2}\oint_{\partial\Omega}(x\,dy-y\,dx)$.
On the boundary we integrate $x\,dy-y\,dx$ (not $Q_x-P_y$). $\partial\Omega$ is three pieces (CCW from the origin):
| Piece | Parametrization | $x\,dy-y\,dx$ | Value |
|---|---|---|---|
| C1 arc | $x=\cos t$, $y=\sin t$, $t:0\to\pi/2$ | $\cos t\cdot\cos t\,dt-\sin t\cdot(-\sin t)\,dt=dt$ | $\int_0^{\pi/2}dt=\pi/2$ |
| C2 $x$-axis | $(1,0)\to(0,0)$, $y=0$, $dy=0$ | $x\,dy=0$ and $y=0\Rightarrow -y\,dx=0$ | $0$ |
| C3 $y$-axis | $(0,0)\to(0,1)$, $x=0$, $dx=0$ | $-y\,dx=0$ and $x=0\Rightarrow x\,dy=0$ | $0$ |
Check: $\frac{1}{4}\pi R^2$ with $R=1$. (Direct $\iint$ in polar also gives $\pi/4$.)
Example (Archetype A — given $\oint x\,dy$ on the unit square):
Given: $\Omega=[0,1]^2$, $\partial\Omega$ counterclockwise. Integrand given: $\oint_{\partial\Omega} x\,dy$ only $\Rightarrow$ $P=0$, $Q=x$.
Asked for: that circulation. On the boundary: integrate $0\,dx+x\,dy=x\,dy$ along each edge.
Route B (interior): $Q_x=1$, $P_y=0$ $\Rightarrow$ $Q_x-P_y=1$.
Route A (boundary): bottom and top have $dy=0$ $\Rightarrow 0$; right edge $x=1$, $\int_0^1 1\,dy=1$; left edge $x=0$ $\Rightarrow 0$. Sum $=1$.
Left panel (swirl plot below): $\displaystyle\oint_{\partial\Omega}(P\,dx+Q\,dy)$ is an integral along the boundary curve. Gold arrows on $\partial\Omega$ show which way you walk (CCW; $\Omega$ on the left) — that orientation was always there; earlier they were orange and easy to confuse with $P$. Along each tiny step $d\mathbf{r}=(dx,dy)$ you add $P\,dx+Q\,dy$ using the field at that boundary point (muted arrows show $\mathbf{v}=(P,Q)$ in $\Omega$). At the probe $(x_0,y_0)$ only: orange horizontal $=P$ (coefficient of $dx$), red vertical $=Q$ (coefficient of $dy$), dark diagonal $=\mathbf{v}$ — a decomposition at one interior point, not the walk direction. On the circular arc, gold ticks can look parallel to $\mathbf{v}=(-y,x)$ by geometry; they are still traversal, not $P$. Right panel: $Q_x-P_y$ mosaic (here constant $2$).
Example (Archetype A — given $\mathbf{v}=(-y,x)$ on the quarter disk):
Given: $\Omega=\{x^2+y^2\le 1,\ x\ge 0,\ y\ge 0\}$ (same as the plot above). Vector field $\mathbf{v}(x,y)=(-y,x)$, so $P=-y$, $Q=x$. $\partial\Omega$ counterclockwise, $\Omega$ on the left.
Asked for: circulation $\displaystyle\oint_{\partial\Omega}(-y\,dx+x\,dy)=\oint_{\partial\Omega}(P\,dx+Q\,dy)$.
What you integrate on the boundary: $-y\,dx+x\,dy$ at each point on the curve (field along the path). Not $Q_x-P_y$ on $\partial\Omega$.
Route A — boundary ($\partial\Omega$ in three pieces, CCW from the origin):
| Piece | Parametrization | $-y\,dx+x\,dy$ | Value |
|---|---|---|---|
| C1 arc | $x=\cos t$, $y=\sin t$, $t:0\to\pi/2$ | $-y\,dx+x\,dy=\sin^2 t\,dt+\cos^2 t\,dt=dt$ | $\int_0^{\pi/2}dt=\pi/2$ |
| C2 $x$-axis | $(1,0)\to(0,0)$, $y=0$, $dy=0$ | $-y\,dx=0$ and $x\,dy=0$ | $0$ |
| C3 $y$-axis | $(0,0)\to(0,1)$, $x=0$, $dx=0$ | $-y\,dx=0$ and $x\,dy=0$ | $0$ |
Sum (Route A): $\pi/2+0+0=\pi/2$. Only the arc contributes; on C2, $dy=0$ and $y=0$ kill both terms; on C3, $dx=0$ and $x=0$ kill both terms.
Route B — interior (Green): same $P,Q$. $Q_x=1$, $P_y=-1$ $\Rightarrow$ $Q_x-P_y=2$.
Why prefer Route B here: $Q_x-P_y$ is constant — one double integral. Route A is correct but needs three parametrizations.
Check: plot caption $\approx\pi/2$; Routes A and B agree.
Exam-style (Archetype A — when the answer is $0$):
Given: $\Omega=[0,2]\times[0,1]$, $\partial\Omega$ counterclockwise. $\displaystyle\oint_{\partial\Omega}(y^2\,dx+2xy\,dy)$.
Read off: $P=y^2$, $Q=2xy$. On the boundary you would integrate $y^2\,dx+2xy\,dy$ along four edges.
Interior (Green): $P_y=2y$, $Q_x=2y$ $\Rightarrow$ $Q_x-P_y=0$.
Meaning: no swirl inside $\Rightarrow$ no net circulation. (If the problem had $Q=2xy+x$ instead, then $Q_x-P_y=1$ and $\oint=\mathrm{Area}(\Omega)=2$.)
Connection to §10.1.9 (conservative fields)
If $\mathbf{v}=(P,Q)=\nabla\phi$, then $P_y=Q_x$ (no swirl). Green’s left side is $0$, so $\oint_{\partial\Omega}\mathbf{v}\cdot d\mathbf{r}=0$ on any closed boundary in a region without holes — consistent with $\int_A^B \mathbf{v}\cdot d\mathbf{r}=\phi(B)-\phi(A)$ on open curves only.
11.4.1 Poincaré lemma (2D) #
Idea
Green converts $\iint_\Omega$ ↔ $\oint_{\partial\Omega}$ for every smooth $(P,Q)$. Poincaré asks when $\mathbf{v}=(P,Q)$ is a gradient on a simply connected $\Omega$ (no holes): conservative $\Leftrightarrow$ $P_y=Q_x$ $\Leftrightarrow$ all closed $\oint=0$.
On an annulus, $P_y=Q_x$ can hold yet $\oint$ around the inner circle is $\neq 0$ — no global $\phi$ on all of $\Omega$ (see example below). Topology: annulus is connected but not simply connected (§14.6).
| Green | Poincaré (2D) | |
|---|---|---|
| Purpose | Convert $\iint_\Omega$ ↔ $\oint_{\partial\Omega}$ | Equivalence: conservative ↔ $P_y=Q_x$ ↔ all closed $\oint=0$ |
| Needs $P_y=Q_x$? | No | That is the derivative test |
| Holes | $\partial\Omega$ has outer + inner curves; apply to each | $P_y=Q_x$ on annulus $\not\Rightarrow$ global potential |
On a simply connected bounded $C^{1}_{\mathrm{pw}}$ domain $\Omega$, for $\mathbf{v}=(P,Q)\in C^1(\bar\Omega;\mathbb{R}^2)$, the following are equivalent:
- $\mathbf{v}$ is conservative ($\mathbf{v}=\nabla\phi$ on $\Omega$).
- $\displaystyle\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$ on $\Omega$.
- $\displaystyle\oint_\gamma \mathbf{v}\cdot d\mathbf{r}=0$ for every closed curve $\gamma$ in $\Omega$.
When not to use
- Domain has a hole (annulus): $P_y=Q_x$ may hold but $\oint$ around the hole can be $\neq 0$ — no global potential on all of $\Omega$.
Example (annulus — Poincaré fails):
On $\Omega=\{0.5\le r\le 1.5\}$ (see annulus plot in §11.3.2), take
Away from the origin, $P_y=Q_x$ (you can check). But for the circle $x^2+y^2=1$ traversed counterclockwise,
So $\mathbf{v}$ is not conservative on $\Omega$ despite $P_y=Q_x$ — the hole at the origin blocks a global potential. Use both boundaries of $\Omega$ if you apply Green on an annulus.
The rotation field below is tangential on circles; right panel: $Q_x-P_y\approx 0$ (flat tiles) even though $P_y=Q_x$ away from the origin. Caption: $\oint_{\partial\Omega}$ can still be nonzero on the outer boundary — the hole blocks a global potential (Poincaré fails).
11.5 Surface Integrals #
Idea
Scalar surface integral: integrate a density $f$ over a curved surface (total mass of a thin shell). Flux integral: how much of a vector field passes through the surface — think of flow through an oriented screen.
At a point on the surface, $\partial_u\Phi$ and $\partial_v\Phi$ span the tangent plane. The vector $\partial_u\Phi\times\partial_v\Phi$ is normal to the patch, and its length $\|\partial_u\Phi\times\partial_v\Phi\|$ is the area of the parallelogram spanned by the parameter directions — that is why $dS$ carries $\|\partial_u\Phi\times\partial_v\Phi\|\,du\,dv$, not $\|\Phi_u\|\,\|\Phi_v\|$. Flux uses the same area element times the component of $\mathbf{K}$ along the chosen normal.
Parametrization $\Phi:B\to\mathbb{R}^3$, $B\subset\mathbb{R}^2$.
Scalar (first kind) #
Vector / flux (second kind) #
Orientation: $d\mathbf{o}=\mathbf{n}\,dS$ with $\mathbf{n}$ unit normal; the cross product $\partial_u\Phi\times\partial_v\Phi$ gives the correct orientation if $\mathbf{n}$ points the way you want (outward for closed surfaces enclosing volume).
Recipe (ZSF-style)
- Parametrize surface $\Phi(u,v)$.
- Compute $\Phi_u$, $\Phi_v$, then $\mathbf{N}=\Phi_u\times\Phi_v$.
- Check normal direction (flip $(u,v)$ order if needed).
- Substitute into integral.
Typical mistake Using $\|\Phi_u\|\,\|\Phi_v\|$ instead of $\|\Phi_u\times\Phi_v\|$ — wrong except for orthogonal parameter lines in special cases.
Before you integrate: scalar vs flux
Scalar (first kind) $\displaystyle\iint_A f\,dS$ — add up $f$ over surface area (mass if $f=\rho$). Integrand uses $\|\partial_u\Phi\times\partial_v\Phi\|$ only (no direction).
Flux (second kind) $\displaystyle\iint_A \mathbf{K}\cdot d\mathbf{o}$ — how much of $\mathbf{K}$ crosses the oriented surface. Same patch area vector $\partial_u\Phi\times\partial_v\Phi$; integrand is $\mathbf{K}\cdot(\partial_u\Phi\times\partial_v\Phi)$ (flow along the chosen normal). Wrong orientation $\Rightarrow$ overall sign flip.
Graph $z=g(x,y)$ over $D$: $\Phi(x,y)=(x,y,g(x,y))$, $\Phi_x\times\Phi_y=(-g_x,-g_y,1)$ (upward normal if $g$ is a graph over $D$).
Worked order (do not skip): (1) $\Phi$ and parameter bounds $B$ → (2) $\Phi_u$, $\Phi_v$, $\mathbf{N}=\Phi_u\times\Phi_v$ → (3) build $f\,\|\mathbf{N}\|$ or $\mathbf{K}\cdot\mathbf{N}$ → (4) write $\int\int_B(\cdots)\,du\,dv$ → (5) inner integral, then outer (or change to polar/cylindrical on $D$).
Example A (scalar — area of a hemisphere)
Given: upper hemisphere $x^2+y^2+z^2=R^2$, $z\ge 0$. Find surface area ($f\equiv 1$).
Step 1 — Parametrize ($u=\vartheta$, $v=\varphi$, see §11.3.3):
Step 2 — Area element (standard spherical patch; same $r^2\sin\vartheta$ as the §11.3.3 Jacobian plot):
Step 3 — Set up the surface integral ($f\equiv 1$):
Step 4 — Inner integral in $\vartheta$:
Step 5 — Outer integral in $\varphi$:
Check: half of full sphere area $4\pi R^2$.
Example B (flux — dot product not constant on a graph)
Given: graph $z=g(x,y)=x^2+y^2$ over $D=\{x^2+y^2\le 1\}$, $\mathbf{K}(x,y,z)=(0,0,z)$, upward normal.
Step 1 — Graph data: $g_x=2x$, $g_y=2y$, $\Phi(x,y)=(x,y,x^2+y^2)$.
Step 2 — Normal vector (upward on a graph):
Step 3 — Field on the surface (substitute $z=g(x,y)$):
Step 4 — Integrand $\mathbf{K}\cdot(\Phi_x\times\Phi_y)$:
Step 5 — Flux as a double integral over $D$:
Step 6 — Polar on $D$ ($x=r\cos\varphi$, $y=r\sin\varphi$, $0\le r\le 1$, $0\le\varphi\le 2\pi$, $dx\,dy=r\,dr\,d\varphi$):
Step 7 — Inner integral in $r$:
Step 8 — Outer integral in $\varphi$:
Contrast: if $\mathbf{K}=(0,0,1)$, Step 4 gives $1$ everywhere and flux $=\mathrm{Area}(D)=\pi$ (Example C) — you would miss the $x^2+y^2$ factor.
Example C (exam archetype D10 — horizontal disk)
Given: $\mathbf{K}=(0,0,1)$, surface $z=1$ over $D=\{x^2+y^2\le 1\}$, upward normal.
Step 1 — Parametrize: $\Phi(x,y)=(x,y,1)$, $(x,y)\in D$; here $g\equiv 1$ so $g_x=g_y=0$.
Step 2 — Normal and dot product:
Step 3 — Flux over $D$:
Step 4 — Area of the unit disk (polar):
Check: constant upward flow through a unit disk — flux equals area. When to use: $\mathbf{K}\parallel$ normal and a flat graph — fewest steps on exams.
Example D (sphere patch — area element)
Given: spherical patch $r=R$, $\vartheta\in[\vartheta_1,\vartheta_2]$, $\varphi\in[\varphi_1,\varphi_2]$ (same $\Phi$ as Example A).
Step 1 — Substitute into the scalar template ($f$ evaluated at $\Phi(\vartheta,\varphi)$):
Mini case: $R=1$, $\vartheta,\varphi\in[0,\pi/2]$ (first-octant patch), $f\equiv 1$.
Step 2 — Set up:
Step 3 — Inner integral:
Step 4 — Outer integral:
Link: in the §11.3.3 spherical patch plot, $r^2\sin\vartheta$ is exactly $\|\Phi_\vartheta\times\Phi_\varphi\|$ on the shell at $r=r_0$.
11.6 Volume Integrals #
Idea
$\iiint_V f\,dV$ sums $f$ over a solid region in $\mathbb{R}^3$ — mass if $f=\rho$ is density, volume if $f\equiv 1$. This is the same Riemann/Jordan picture as §11.1, now in three variables.
Compute by
- Fubini on a normal region in $\mathbb{R}^3$ (§11.2.1): fix $(x,y)\in D$, integrate $z$ between graphs $\varphi_1\le z\le\varphi_2$, then integrate over $D$.
- Substitution when $V$ is a ball, cylinder, or sector — polar / cylindrical / spherical (§11.3) with the Jacobian factor.
Do not mix up surface integrals (§11.5, 2D parameter domain) with volume integrals (solid $V\subset\mathbb{R}^3$).
Typical mistake Setting up a volume integral without a clear order of integration — draw the projection onto the $(x,y)$-plane first, then fix $z$-bounds.
Before you integrate: volume in $\mathbb{R}^3$
Volume integral $\displaystyle\iiint_V f\,dV$ sums $f$ over a solid $V\subset\mathbb{R}^3$ (filled region). This is not a §11.5 integral (those live on a 2D surface $\partial V$).
- $f\equiv 1$ $\Rightarrow$ volume of $V$.
- $f=\rho$ (density) $\Rightarrow$ mass $M=\iiint_V \rho\,dV$.
Route 1 — Fubini (§11.2.1): if $V=\{(x,y,z):(x,y)\in D,\ \varphi_1(x,y)\le z\le\varphi_2(x,y)\}$,
Route 2 — substitution (§11.3): ball / cylinder / sector $\Rightarrow$ cylindrical $dV=r\,dr\,d\varphi\,dz$ or spherical $dV=r^2\sin\vartheta\,dr\,d\vartheta\,d\varphi$.
Worked order (do not skip): describe $V$ and projection $\Rightarrow$ pick route $\Rightarrow$ write all bounds $\Rightarrow$ inner integral $\Rightarrow$ middle $\Rightarrow$ outer $\Rightarrow$ check (symmetry, known volume table in §11.3.1).
Example A (volume — cylinder via Fubini, then cylindrical)
Given: solid cylinder $x^2+y^2\le 1$, $0\le z\le h$. Find volume ($f\equiv 1$).
Step 1 — Normal region (§11.2.1): disk $D=\{x^2+y^2\le 1\}$ in the $(x,y)$-plane, $0\le z\le h$.
Step 2 — Integrate in $z$ first:
Step 3 — Polar on $D$ ($x=r\cos\varphi$, $y=r\sin\varphi$, $0\le r\le 1$, $0\le\varphi\le 2\pi$, $dA=r\,dr\,d\varphi$):
Step 4 — Inner integral in $r$:
Step 5 — Outer integral in $\varphi$:
Result: $V=h\cdot\pi=\pi h$.
Check: matches cylinder volume $\pi R^2 h$ from §11.3.1 with $R=1$. See the §11.3.3 cylindrical coordinates plot.
Example B (mass on unit cube — exam-style Fubini)
Given: $V=[0,1]^3$, $\rho(x,y,z)=xyz$. Find mass $M$ (same setup as §11.9 exercise 11).
Step 1 — Set up:
Step 2 — Inner integral in $z$ ($x,y$ fixed):
Step 3 — Middle integral in $y$:
Step 4 — Outer integral in $x$:
Check: on a box, $\iiint xyz\,dV=\bigl(\int_0^1 x\,dx\bigr)\bigl(\int_0^1 y\,dy\bigr)\bigl(\int_0^1 z\,dz\bigr)=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}$.
Example C (mass on unit ball — spherical substitution)
Given: unit ball $V=\{x^2+y^2+z^2\le 1\}$, $\rho(x,y,z)=1-x^2-y^2-z^2$.
Step 1 — Why spherical: $\rho$ depends on $r^2=x^2+y^2+z^2$ only; $V$ is a ball $0\le r\le 1$.
Step 2 — Substitute ($\rho=1-r^2$, $dV=r^2\sin\vartheta\,dr\,d\vartheta\,d\varphi$, bounds $0\le r\le 1$, $0\le\vartheta\le\pi$, $0\le\varphi\le 2\pi$):
Step 3 — Factor (integrand is a product of a function of $r$, $\sin\vartheta$, and $1$):
Step 4 — $r$-integral:
Step 5 — $\vartheta$-integral:
Step 6 — $\varphi$-integral:
Step 7 — Combine:
Check: $\rho\ge 0$ on $V$ and $M>0$. For volume of the same ball ($f\equiv 1$), use the stepped spherical example in §11.3.3 — do not repeat that calculation here.
11.7 Center of Mass (Schwerpunkt) #
Idea
The center of mass is the balance point: each coordinate $s_{x_i}$ is the average position weighted by density $\rho$. You always need the total mass $M=\iiint_K \rho\,dV$ in the denominator — same volume machinery as §11.6.
Formulas
For a solid $K\subset\mathbb{R}^n$ with density $\rho$ (often $\rho\equiv 1$):
Symmetry shortcuts
- If $K$ is symmetric under $x\mapsto -x$ and $\rho$ is even in $x$, then $s_x=0$ without integrating. Similarly for $y,z$.
- Uniform density on a body symmetric about the $z$-axis $\Rightarrow$ $s_x=s_y=0$; only $s_z$ may need a calculation.
Typical mistake
- Using $\iiint x_i\,\rho\,dV$ as the center of mass — forgets dividing by $M$.
- Ignoring $\rho$ in $M$ while integrating $x_i\rho$ in the numerator (inconsistent density).
Example:
Uniform hemisphere $x^2+y^2+z^2\le R^2$, $z\ge 0$. By symmetry $s_x=s_y=0$. For $s_z$, use spherical coordinates — result $s_z=\frac{3}{8}R$ (center of mass lies on $z$-axis above origin).
11.8 Gauss's Theorem (Divergence Theorem) #
Idea
Why only the boundary appears (box partition) — same logic as Green (§11.4): partition $V$ into small boxes; on each box, outward flux through its faces matches $\mathrm{div}\,\mathbf{v}$ on that box. Interior faces cancel (outward from one box = inward into the neighbor). Only $\partial V$ remains — volume integral of $\mathrm{div}\,\mathbf{v}$ becomes one flux integral on the closed boundary.
What $\mathrm{div}\,\mathbf{v}$ means — local source rate (fluid created per unit volume). Positive div → spreading; negative div → sink. Total creation inside $V$ = net outward flux through $\partial V$.
When to use — Closed surface + solid $V$, field smooth on $\bar V$ → try $\iiint_V \mathrm{div}\,\mathbf{v}\,dV$ instead of parametrizing $\partial V$. Same flux as §11.5; Gauss is the shortcut when the surface encloses a volume.
Definition (divergence)
In $\mathbb{R}^3$: $\mathrm{div}\,(v_1,v_2,v_3)=\partial_x v_1+\partial_y v_2+\partial_z v_3$.
Gauss Let $V\subset\mathbb{R}^3$ be bounded with $C^{1}_{\mathrm{pw}}$ boundary $\partial V$, outward unit normal $\mathbf{n}$. For $\mathbf{v}\in C^1(\bar V;\mathbb{R}^3)$:
When to use
- Closed surface + field smooth inside → try div and volume integral (often easier than parametrizing $\partial V$).
- Compare with §11.5 flux: same integral, Gauss is the shortcut when $V$ is solid and closed.
Synthesis: operator identities, Stokes, and when to pick Green vs Stokes vs Gauss — Ch 13 §13.3, §13.5.
Typical mistake Wrong orientation (inward normal) — pick outward $\mathbf{n}$ for the standard Gauss formula.
Useful identities
- $\mathrm{div}(\nabla\times\mathbf{K})=0$.
- $\mathrm{div}(f\mathbf{K})=\nabla f\cdot\mathbf{K}+f\,\mathrm{div}\,\mathbf{K}$.
Example (flux of radial field through sphere):
$\mathbf{v}(x,y,z)=(x,y,z)$, $V=$ ball radius $R$. Then $\mathrm{div}\,\mathbf{v}=3$.
Direct check: on the sphere, $\mathbf{n}=\mathbf{x}/R$, $\mathbf{v}\cdot\mathbf{n}=R$, area $4\pi R^2$, flux $R\cdot 4\pi R^2=4\pi R^3$.
11.9 Integration in $\mathbb{R}^n$ — Exercises #
Use §11.1–11.8 and Chapter 10 line integrals where needed.
A) Fubini and normal regions #
- $\displaystyle\int_0^1\int_0^{x^2} (x+y)\,dy\,dx$
- Area of $\{(x,y):0\le y\le 1,\ y\le x\le 2-y\}$
- $\displaystyle\iiint_V z\,dV$ where $V$ is the box $[0,1]^3$
B) Jacobian (polar / cylindrical / spherical) #
- $\displaystyle\iint_{x^2+y^2\le 4} (x^2+y^2)\,dA$
- $\displaystyle\int_0^{2\pi}\int_0^1\int_0^2 r\,dz\,dr\,d\varphi$ (cylinder setup — evaluate)
- Volume of $\{(x,y,z):x^2+y^2+z^2\le 9,\ z\ge 0\}$ (hemisphere)
C) Green and conservative fields #
- $\displaystyle\oint_\gamma (y\,dx+2x\,dy)$ where $\gamma$ is the unit circle counterclockwise
- Is $\mathbf{v}=(2xy+y^2,\ x^2+2xy)$ conservative on $\mathbb{R}^2$? If yes, find a potential.
D) Surface integrals #
- $\displaystyle\iint_A z\,dS$ where $A$ is the plane $z=1$ over $x^2+y^2\le 1$
- Flux of $\mathbf{K}=(0,0,1)$ through $z=1$, $x^2+y^2\le 1$ (upward normal)
E) Volume and center of mass #
- Mass of unit cube $[0,1]^3$ with $\rho(x,y,z)=xyz$
- $s_z$ for uniform solid cylinder $x^2+y^2\le 1$, $0\le z\le 2$
F) Gauss #
- $\displaystyle\iint_{\partial V}\mathbf{v}\cdot\mathbf{n}\,dS$ for $\mathbf{v}=(x^2,y,z)$, $V=$ unit cube $[0,1]^3$
- $\mathrm{div}$ of $(yz,xz,xy)$ and flux through sphere radius $1$ centered at origin
Quick decision checklist #
- Box or normal region? → Fubini with explicit bounds (§11.1.3, §11.2).
- Disk, annulus, ball, cylinder? → polar / cylindrical / spherical (§11.3).
- $\oint P\,dx+Q\,dy$ on closed curve in $\mathbb{R}^2$? → Green or check $P_y=Q_x$ (§11.4, §10.1.9).
- Flux through a surface? → parametrize (§11.5); if closed surface of a solid → Gauss (§11.8).
- Mass / center of mass? → $\iiint \rho\,dV$ and symmetry (§11.6–11.7).
Answers (final results) #
A) 1. $\frac{7}{30}$ — 2. $1$ — 3. $\frac{1}{2}$
B) 4. $8\pi$ — 5. $4\pi$ — 6. $18\pi$
C) 7. $\pi$ — 8. yes, $\phi(x,y)=x^2y+xy^2+C$
D) 9. $\pi$ — 10. $\pi$
E) 11. $\frac{1}{8}$ — 12. $s_z=1$
F) 13. $3$ — 14. $\mathrm{div}=0$, flux $0$
Worked solutions (selected) #
A1
B4 Polar: $\int_0^{2\pi}\int_0^2 r^2\cdot r\,dr\,d\varphi=2\pi\cdot\frac{16}{4}=8\pi$.
C7 Green: $\partial Q/\partial x-\partial P/\partial y=2-1=1$, area of unit disk $=\pi$.
C8 $P=2xy+y^2$, $Q=x^2+2xy$. Then $P_y=2x+2y$, $Q_x=2x+2y$ — equal on $\mathbb{R}^2$, so conservative. Find $\phi$:
So $\phi(x,y)=x^2y+xy^2+C$. Verify: $\phi_x=2xy+y^2=P$, $\phi_y=x^2+2xy=Q$.
D10 $\mathbf{K}=(0,0,1)$ on $z=1$, $x^2+y^2\le 1$, upward. $\Phi=(x,y,1)$, $\Phi_x\times\Phi_y=(0,0,1)$, $\mathbf{K}\cdot(\Phi_x\times\Phi_y)=1$, flux $=\iint_{x^2+y^2\le 1}1\,dx\,dy=\pi$.
F13 $\mathrm{div}\,\mathbf{v}=2x+1+1=2x+2$,
F14 $\partial_x(yz)+\partial_y(xz)+\partial_z(xy)=0+0+0$, flux $0$ by Gauss.
Chapter 12 — Potential fields (Potenzialfelder) #
A potential field is a vector field that is a gradient: an “antiderivative” in $\mathbb{R}^2$ or $\mathbb{R}^3$, so line integrals reduce to endpoint differences.
Story of the chapter: if $\mathbf{v}=\nabla\phi$ on $\Omega$, work along any path in $\Omega$ depends only on endpoints, not on the route. Finding $\phi$ is an integration problem. Topology (simply connected vs holes) decides whether “curl-free” ($P_y=Q_x$) actually implies a global $\phi$.
Prerequisites: §10.1.9 (exam recipe), §11.4.1 (holes), §11.4 connection to conservative fields.
Where to start
Operational exam steps (test $P_y=Q_x$, find $\phi$, verify, use $\phi(B)-\phi(A)$): §10.1.9.
This chapter explains why that works, when it fails, and a second method (path integral) to find $\phi$.
12.0 What a potential field buys you #
Definition: $\phi\in C^1(\Omega)$ is a potential of $\mathbf{v}=(P,Q)$ on $\Omega$ if $\mathbf{v}=\nabla\phi$, i.e. $\phi_x=P$ and $\phi_y=Q$ on $\Omega$.
Before you use $\phi(B)-\phi(A)$
Payoff: if $\mathbf{v}=\nabla\phi$ on $\Omega$ and your path $\gamma$ from $A$ to $B$ lies in $\Omega$, then
(path independent — same endpoints, same value).
On a simply connected domain $\Omega$ (no holes; see §14.6), for $\mathbf{v}\in C^1(\Omega)$ these are equivalent:
- $\mathbf{v}=\nabla\phi$ on $\Omega$ for some $\phi$;
- $P_y=Q_x$ everywhere on $\Omega$;
- $\displaystyle\oint_\gamma \mathbf{v}\cdot d\mathbf{r}=0$ for every closed curve $\gamma$ in $\Omega$.
Not equivalent if $\Omega$ has a hole: $P_y=Q_x$ on an annulus need not give a global $\phi$ — see §12.3.
Rule: never use $\phi(B)-\phi(A)$ without verifying $\nabla\phi=\mathbf{v}$ (§10.1.9, Step 3).
$n$-dimensional integrability: on simply connected $\Omega\subset\mathbb{R}^n$, $\mathbf{v}$ is a gradient iff $\partial v_i/\partial x_j=\partial v_j/\partial x_i$ for all $i\ne j$.
12.1 Method A — partial integration (see §10.1.9) #
Full step table: §10.1.9 (test $P_y=Q_x$ $\Rightarrow$ $\phi=\int P\,dx+h(y)$ $\Rightarrow$ match $Q$ $\Rightarrow$ verify).
Example A (find $\phi$ by integrating $P$ — same field as §11.9 C8)
Given: $\mathbf{v}(x,y)=(2xy+y^2,\ x^2+2xy)$ on $\mathbb{R}^2$.
Step 1 — Conservative test:
Step 2 — Integrate $P$ w.r.t. $x$ ($y$ fixed):
Step 3 — Match $Q$:
Step 4 — Verify:
So $\phi(x,y)=x^2y+xy^2+C$.
Step 5 — Use potential (optional): from $(0,0)$ to $(1,1)$,
12.2 Method B — path integral along a broken line #
If Method A is awkward, fix a base point $(x_0,y_0)$ and integrate along horizontal then vertical (or the reverse, staying in $\Omega$):
Example B (path integral route — exercise B1 field)
Given: $\mathbf{v}=(e^x\cos y,\,-e^x\sin y)$ on $\mathbb{R}^2$.
Step 1 — Test: $P_y=-e^x\sin y$, $Q_x=-e^x\sin y$ $\Rightarrow$ $P_y=Q_x$.
Step 2 — Base point $(x_0,y_0)=(0,0)$. Target $(x,y)$.
Step 3 — Leg 1: $(0,0)\to(x,0)$ along $y=0$, $dy=0$:
Step 4 — Leg 2: $(x,0)\to(x,y)$ along $x$ fixed, $dx=0$:
Step 5 — Combine:
With $\phi(0,0)=0$: $\phi(x,y)=e^x\cos y-1$ (any $+C$ is fine for endpoint differences).
Step 6 — Verify:
12.3 When $P_y=Q_x$ is not enough (holes) #
Simply connected $\Omega$: every closed curve in $\Omega$ can be shrunk to a point inside $\Omega$ (no holes). For $\mathbf{v}\in C^1(\Omega)$,
With a hole (e.g. annulus around the origin): $P_y=Q_x$ can still hold on $\Omega$, but a loop around the hole may have $\oint\neq 0$, so there is no single $\phi$ on all of $\Omega$ with $\nabla\phi=\mathbf{v}$.
Three situations (do not mix them up)
| Situation | Domain $\Omega$ | $P_y=Q_x$? | Global $\phi$ on $\Omega$? | What went wrong |
|---|---|---|---|---|
| A — simply connected | $\mathbb{R}^2$ (no hole) | yes | yes | nothing — use §12.1 / §10.1.9 |
| B — hole | annulus $0.5\le r\le 1.5$ | yes | no | topology: circulation around the hole |
| C — swirl on plane | $\mathbb{R}^2$ | no | no | curl test already fails (not a hole story) |
Examples A and B below are the pair you want for hole vs no hole. Example C is only there so you do not confuse “$\oint\neq 0$” with “must be a hole”.
Example A — no hole: $P_y=Q_x$ $\Rightarrow$ potential exists
Given: $\mathbf{v}(x,y)=(2x,2y)$ on $\Omega=\mathbb{R}^2$ (simply connected).
Asked: Is $\mathbf{v}=\nabla\phi$? If yes, find $\phi$ and evaluate $\displaystyle\int_{(0,0)}^{(1,2)}\mathbf{v}\cdot d\mathbf{r}$.
Step 1 — Curl test: $P=2x$, $Q=2y$.
On a simply connected domain, this is enough to expect a potential.
Step 2 — Find $\phi$: $\displaystyle\phi(x,y)=\int 2x\,dx=x^2+h(y)$, then $\phi_y=h'(y)=2y$ so $h(y)=y^2+C$.
Step 3 — Verify: $\phi_x=2x=P$, $\phi_y=2y=Q$ $\Rightarrow$ $\nabla\phi=\mathbf{v}$ on $\mathbb{R}^2$.
Step 4 — Endpoint formula (legal now):
Step 5 — Closed loop check (optional): any closed $\gamma$ in $\mathbb{R}^2$ gives $\oint \mathbf{v}\cdot d\mathbf{r}=0$ (e.g. $\phi$ returns to the same value). No hole $\Rightarrow$ no hidden circulation.
(Same field as §10.1.9.)
Example B — hole: $P_y=Q_x$ but no global potential
Given: annulus $\Omega=\{0.5\le r\le 1.5\}$ (hole at the origin; plot in §11.3.2),
Asked: Does $P_y=Q_x$ on $\Omega$? Is $\mathbf{v}=\nabla\phi$ on all of $\Omega$?
Step 1 — Curl test on $\Omega$ (away from the origin):
Step 2 — Do not stop here. On an annulus, $P_y=Q_x$ does not guarantee a global $\phi$. Check a loop around the hole.
Step 3 — Circulation on $x^2+y^2=1$ (CCW, $t\in[0,2\pi]$):
Step 4 — Conclusion: $\mathbf{v}$ is not conservative on $\Omega$ — no $\phi\in C^1(\Omega)$ with $\nabla\phi=\mathbf{v}$ on the whole annulus.
Step 5 — Why students get trapped: polar angle $\theta=\arctan(y/x)$ looks like $\partial\theta/\partial x=-y/r^2$, $\partial\theta/\partial y=x/r^2$, but $\theta$ is not single-valued on $\Omega$ (one lap adds $2\pi$). That is the hole, not a algebra slip.
Visual: §11.4.1 — green plot, rotation field on the annulus.
Example C — no hole, but curl test fails (not Example B)
Given: $\mathbf{v}=(-y,x)$ on $\mathbb{R}^2$.
Step 1: $P_y=-1$, $Q_x=1$ $\Rightarrow$ $P_y\neq Q_x$.
Step 2: Stop — not conservative on $\mathbb{R}^2$; do not hunt for $\phi$, and do not blame a hole (there is none). Optional check: $\oint_{\text{unit circle}}(-y\,dx+x\,dy)=2\pi\neq 0$.
This is §11.4 Archetype A swirl on a disk — Green gives nonzero circulation because $Q_x-P_y=2$, not because of topology.
12.4 Quick decision checklist (exam) #
- Closed curve in $\mathbb{R}^2$? $\Rightarrow$ try $\oint(P\,dx+Q\,dy)$ or Green (§11.4); use $\phi(B)-\phi(A)$ only if you know $\mathbf{v}$ is conservative on the region.
- Open curve, simply connected region? $\Rightarrow$ test $P_y=Q_x$ $\Rightarrow$ find $\phi$ (§10.1.9, §12.1, or §12.2) $\Rightarrow$ verify $\Rightarrow$ $\int_A^B\mathbf{v}\cdot d\mathbf{r}=\phi(B)-\phi(A)$.
- Hole / annulus? $\Rightarrow$ do not assume a global $\phi$ from $P_y=Q_x$ alone; see §11.4.1.
12.5 Exercises and answers #
A1 On the annulus $\{0.5\le r\le 1.5\}$, is $\mathbf{v}=\bigl(-\tfrac{y}{x^2+y^2},\,\tfrac{x}{x^2+y^2}\bigr)$ a gradient?
A2 On $\mathbb{R}^2$, is $\mathbf{v}=(2x,2y)$ a gradient?
B1 Find a potential for $\mathbf{v}=(e^x\cos y,\,-e^x\sin y)$ on $\mathbb{R}^2$.
B2 Using a potential, evaluate $\displaystyle\int_{(0,0)}^{(1,1)}(2xy+y^2,\,x^2+2xy)\cdot d\mathbf{r}$.
Answers
- A1: No. $P_y=Q_x$ on the annulus, but $\oint=2\pi\neq 0$ on a circle around the hole (§12.3, Example B).
- A2: Yes. $\phi=x^2+y^2+C$ (§12.3, Example A).
- B1: $\phi(x,y)=e^x\cos y+C$ (full path-integral solution in §12.2).
- B2: $\phi=x^2y+xy^2+C$ gives $\phi(1,1)-\phi(0,0)=2$ (§12.1, Example A).
Chapter 13 — Vector analysis (Vektoranalysis) #
grad, div, rot, operator identities, and a map of Green / Stokes / Gauss — when to use which.
Story of the chapter: Chapter 11 shows how to compute flux and volume integrals. Here you learn what $\mathrm{grad}$, $\mathrm{div}$, and $\mathrm{rot}$ mean, and how the three big theorems are one ladder: interior derivative $\leftrightarrow$ boundary integral. Stokes in 3D is the piece Green does not cover alone.
Prerequisites: Chapter 11 (§11.4 Green, §11.5 flux, §11.8 Gauss); Chapter 12 ($\mathbf{v}=\nabla\phi$, curl-free fields).
Where to start
Compute a flux or volume integral (parametrize, Fubini, Jacobian): §11.5, §11.6, §11.8.
This chapter — interpret $\mathrm{grad}/\mathrm{div}/\mathrm{rot}$, choose Green vs Stokes vs Gauss, and work Stokes examples Ch 11 does not spell out.
13.0 FTC ladder (Green / Stokes / Gauss) #
Same idea as the Ch 11 theorem ladder: partition the domain into small pieces; interior contributions cancel; only the boundary remains.
| Theorem | Region | Interior (divergence / curl) | Boundary | Operator form |
|---|---|---|---|---|
| Green (2D) | plane region $\Omega$ | $\displaystyle\iint_\Omega (Q_x-P_y)\,dA$ | $\displaystyle\oint_{\partial\Omega}(P\,dx+Q\,dy)$ | $Q_x-P_y=(\mathrm{rot}\,\mathbf{v})_z$ for $\mathbf{v}=(P,Q,0)$ |
| Stokes (3D) | oriented surface $S$ | $\displaystyle\iint_S (\mathrm{rot}\,\mathbf{F})\cdot d\mathbf{S}$ | $\displaystyle\int_{\partial S}\mathbf{F}\cdot d\mathbf{r}$ | circulation on $\partial S$ |
| Gauss (3D) | solid $V$ | $\displaystyle\iiint_V \mathrm{div}\,\mathbf{v}\,dV$ | $\displaystyle\iint_{\partial V}\mathbf{v}\cdot\mathbf{n}\,dS$ | outward flux |
Reading the table: Green is Stokes in the plane (normal $\mathbf{k}$, curl has only a $z$-component). Gauss is the divergence version: sources inside $V$ $\Leftrightarrow$ net outward flux through $\partial V$. Full statements and proofs: §11.4, §13.3, §11.8.
Scalar vs vector on surfaces (ZSF 11.3 / 13)
| Integral | Field type | Measures |
|---|---|---|
| $\displaystyle\iint_A f\,dS$ | scalar $f$ | Mass / area on a shell (§11.5, 1st kind) |
| $\displaystyle\iint_A \mathbf{K}\cdot d\mathbf{o}$ | vector $\mathbf{K}$ | Flux through oriented $A$ (§11.5, 2nd kind); $d\mathbf{o}=\mathbf{n}\,dS$ |
$\mathrm{grad}$ acts on a scalar; $\mathrm{div}$ and $\mathrm{rot}$ act on a vector.
13.1 Operators (grad, div, rot) #
For a scalar $f$ and vector field $\mathbf{F}=(F_1,F_2,F_3)$ in $\mathbb{R}^3$:
(Gradient in $\mathbb{R}^n$: §8.1.5–8.1.6 — Jacobian and gradient.)
Curl mnemonic (ZSF 13.2.2): $\mathrm{rot}\,\mathbf{F}=\nabla\times\mathbf{F}$ is the symbolic determinant
What they measure (geometry)
| Operator | On | Meaning (picture) |
|---|---|---|
| $\mathrm{grad}\,f$ | scalar $f$ | Direction of steepest increase; perpendicular to level surfaces |
| $\mathrm{div}\,\mathbf{F}$ | vector $\mathbf{F}$ | Source rate per unit volume (fluid created); $\mathrm{div}>0$ → spreading |
| $\mathrm{rot}\,\mathbf{F}$ | vector $\mathbf{F}$ | Swirl axis (magnitude = local rotation strength); tangential “spin” |
Mini-examples (compute before an exam)
(i) $\mathbf{F}=(x,y,z)$: $\mathrm{div}\,\mathbf{F}=1+1+1=3$; $\mathrm{rot}\,\mathbf{F}=\mathbf{0}$ (radial field, no swirl).
(ii) $\mathbf{F}=(0,0,z)$: $\mathrm{div}\,\mathbf{F}=0+0+1=1$; $\mathrm{rot}\,\mathbf{F}=\mathbf{0}$.
(iii) $\mathbf{F}=(y,-x,0)$ (rotation in the $xy$-plane): $\mathrm{div}\,\mathbf{F}=0$; $\mathrm{rot}\,\mathbf{F}=(0,0,-2)$.
Exam move — only $\mathrm{rot}\,\mathbf{K}$ is given (ZSF 13.2.2)
Given: $\mathrm{rot}\,\mathbf{K}=(0,0,1)$ on $\mathbb{R}^3$. Find some $\mathbf{K}$ with that curl.
Guess: rotation in the $xy$-plane, e.g. $\mathbf{K}=(-y,x,0)$.
Check:
Sign wrong → try $\mathbf{K}=(y,-x,0)$: $\mathrm{rot}\,\mathbf{K}=(0,0,-2)$. For $(0,0,1)$ scale: $\mathbf{K}=\tfrac{1}{2}(y,-x,0)$ gives $\mathrm{rot}\,\mathbf{K}=(0,0,1)$.
Use: Stokes with known $\mathrm{rot}\,\mathbf{K}$ when $\mathbf{K}$ itself is awkward to integrate on $\partial S$.
13.2 Identities (ZSF 13.2.3) #
Exam use
- $\mathrm{div}(f\mathbf{K})$ — split a flux integral when $\mathbf{v}=f\mathbf{K}$ (see §11.8 product rule).
- $\mathrm{div}(f\,\mathrm{rot}\,\mathbf{K})$ — when flux involves $f$ times a rotation field; reduces to $\nabla f\cdot\mathrm{rot}\,\mathbf{K}$.
- $\mathrm{rot}(\nabla f)=0$ — every gradient field has no swirl; if $\mathbf{v}=\nabla\phi$ then $\mathrm{rot}\,\mathbf{v}=\mathbf{0}$ (Chapter 12). On simply connected $\Omega$, $\mathrm{rot}\,\mathbf{v}=\mathbf{0}$ $\Rightarrow$ path independence (Stokes).
- $\mathrm{div}(\mathrm{rot}\,\mathbf{K})=0$ — swirl fields have no net sources; Gauss on a region with only $\mathrm{rot}\,\mathbf{K}$ often gives flux $0$.
- In $\mathbb{R}^2$, $P_y=Q_x$ for $\mathbf{v}=(P,Q,0)$ $\Leftrightarrow$ $(\mathrm{rot}\,\mathbf{v})_z=0$.
- $\mathrm{div}(\nabla f)=\Delta f$ — Poisson / Laplace context (scalar).
Example ($\mathrm{div}(f\,\mathrm{rot}\,\mathbf{K})$ — no integral)
Given: $f(x,y,z)=x$, $\mathbf{K}=(y,-x,0)$. Find $\mathrm{div}\bigl(f\,\mathrm{rot}\,\mathbf{K}\bigr)$.
Step 1 — Curl: $\mathrm{rot}\,\mathbf{K}=(0,0,-2)$.
Step 2 — Identity: $\mathrm{div}(f\,\mathrm{rot}\,\mathbf{K})=\nabla f\cdot\mathrm{rot}\,\mathbf{K}$.
Step 3 — Dot: $\nabla f=(1,0,0)$, so $\nabla f\cdot\mathrm{rot}\,\mathbf{K}=1\cdot(-2)=-2$.
Check: expand $f\,\mathrm{rot}\,\mathbf{K}=(-2x,0,0)$; $\mathrm{div}=(-2)+0+0=-2$.
13.3 Theorem map (Green / Stokes / Gauss) #
| Question type | First try | Statement / link |
|---|---|---|
| Closed curve in $\mathbb{R}^2$, $(P,Q)$ given | Green | $\displaystyle\oint_{\partial\Omega}(P\,dx+Q\,dy)=\iint_\Omega(Q_x-P_y)\,dA$ — §11.4 |
| Open surface $S$ with boundary, circulation / curl | Stokes | $\displaystyle\iint_S(\mathrm{rot}\,\mathbf{F})\cdot d\mathbf{S}=\int_{\partial S}\mathbf{F}\cdot d\mathbf{r}$ — §13.4 |
| Closed surface of a solid $V$ | Gauss | $\displaystyle\iint_{\partial V}\mathbf{v}\cdot\mathbf{n}\,dS=\iiint_V\mathrm{div}\,\mathbf{v}\,dV$ — §11.8 |
| Open surface, flux only, no shortcut | Parametrize | §11.5 recipe |
Green as 2D Stokes: for $\mathbf{v}=(P,Q,0)$ on a plane region, $Q_x-P_y=(\mathrm{rot}\,\mathbf{v})_z$.
Stokes (classical): oriented surface $S$ with piecewise smooth boundary $\partial S$. Choose $\mathbf{n}$ on $S$; traverse $\partial S$ counterclockwise when viewed from the tip of $\mathbf{n}$ (right-hand rule).
Gauss: outward unit normal $\mathbf{n}$ on closed $\partial V$. Inward normal flips the sign.
Partial boundary (ZSF 13.2.6): Gauss needs a closed $\partial V$. If you only want flux through one piece $S\subset\partial V$ (e.g. a hemisphere), close $V$ mentally, compute $\iiint_V \mathrm{div}\,\mathbf{v}\,dV$, then subtract flux through the other faces of $\partial V\setminus S$ (§13.6).
Orientation — right-hand rule (ZSF)
Flux: $d\mathbf{o}=\mathbf{n}\,dS$; pick $\mathbf{n}$ (outward on closed surfaces, or “through the screen” on open patches).
Stokes / Green: thumb along $\mathbf{n}$ on the surface; fingers curl in the positive direction along $\partial S$ / $\partial\Omega$ (region on your left in 2D).
Wrong $\mathbf{n}$ or wrong traversal direction $\Rightarrow$ overall sign flip.
Area via Green (ZSF 13.2.4 / 13.2.8): $\mathrm{Area}(D)=\displaystyle\iint_D 1\,dA$. With $\mathbf{v}=(0,x,0)$, $(\mathrm{rot}\,\mathbf{v})_z=1$, so $\mathrm{Area}(D)=\oint_{\partial D}\mathbf{v}\cdot d\mathbf{r}$ (CCW). Full steps: §13.4b.
13.4 Stokes — worked example #
Example (unit disk — Stokes matches Green)
Given: $S=$ unit disk in the plane $z=0$, $x^2+y^2\le 1$, normal $\mathbf{n}=\mathbf{k}$ (upward). $\mathbf{F}=(-y,x,0)$.
Asked: Verify Stokes: compare $\displaystyle\iint_S (\mathrm{rot}\,\mathbf{F})\cdot d\mathbf{S}$ and $\displaystyle\oint_{\partial S}\mathbf{F}\cdot d\mathbf{r}$.
Step 1 — Curl:
Step 2 — Surface integral ($d\mathbf{S}=\mathbf{n}\,dS$, $\mathbf{n}=\mathbf{k}$):
Step 3 — Boundary $\partial S$: circle $x=\cos t$, $y=\sin t$, $z=0$, $t\in[0,2\pi]$ (CCW seen from $+\mathbf{k}$):
Step 4 — Conclusion: both sides $=2\pi$; Stokes checks. Same numbers as Green on the disk with $P=-y$, $Q=x$ (§11.4, $Q_x-P_y=2$).
Link to Ch 12: $\mathrm{rot}\,\mathbf{F}\neq\mathbf{0}$ on the disk $\Rightarrow$ $\mathbf{F}$ is not a gradient on any region containing the disk — consistent with $\oint=2\pi\neq 0$.
13.4b Green — area via line integral (ZSF 13.2.8) #
Example (area of unit disk without polar double integral)
Given: $D=\{x^2+y^2\le 1\}\subset\mathbb{R}^2$. Find $\mathrm{Area}(D)$ using Green.
Step 1 — Pick a field with constant curl: $\mathbf{v}=(0,x,0)$ so $P=0$, $Q=x$.
Step 2 — Green: CCW on $\partial D$ gives $\mathrm{Area}(D)=\oint_{\partial D}(0\,dx+x\,dy)$.
Step 3 — Parametrize $\partial D$: $x=\cos t$, $y=\sin t$, $t\in[0,2\pi]$ (CCW, $D$ on the left).
Step 4 — Line integral:
Check: $\pi r^2$ with $r=1$. Same idea as choosing $P=-y$, $Q=0$ ($\oint -y\,dx$).
13.5 Flux: when to parametrize vs Gauss vs Stokes #
Rules
- Closed $\partial V$ + smooth $\mathbf{v}$ on $\bar V$ → compute $\mathrm{div}\,\mathbf{v}$, use §11.8.
- Open surface (hemisphere, graph, patch) → usually §11.5: $\boldsymbol{\Phi}(u,v)$, $\mathbf{N}=\boldsymbol{\Phi}_u\times\boldsymbol{\Phi}_v$, integrate $\mathbf{v}\cdot\mathbf{N}\,du\,dv$.
- Circulation on $\partial S$ with easy $\mathrm{rot}\,\mathbf{F}$ → Stokes (§13.4).
- Need $\phi(B)-\phi(A)$? → Chapter 12, not this chapter.
Flux Kochrezept (ZSF 13.2.7) — exam card
Given: vector field $\mathbf{v}$, surface $S$. Find: $\displaystyle\iint_S \mathbf{v}\cdot\mathbf{n}\,dS$.
| Step | Action |
|---|---|
| 1 | Parametrize $S$: $\boldsymbol{\Phi}(u,v)$ on $[a,b]\times[c,d]$ (§11.5) |
| 2 | $\boldsymbol{\Phi}_u$, $\boldsymbol{\Phi}_v$, $\mathbf{N}=\boldsymbol{\Phi}_u\times\boldsymbol{\Phi}_v$; flip $(u,v)$ if $\mathbf{N}$ points the wrong way |
| 3 | $\displaystyle\iint_S \mathbf{v}\cdot d\mathbf{o}=\int_a^b\int_c^d \mathbf{v}(\boldsymbol{\Phi})\cdot\mathbf{N}\,du\,dv$ |
If $S$ closes a solid and $\mathrm{div}\,\mathbf{v}$ is easy → try Gauss first (§11.8).
Scalar surface area (no vector field): §11.5 Example A (hemisphere area).
13.6 Example — hemisphere flux (exercise C1) #
Example (flux of $(0,0,z)$ through upper hemisphere)
Given: $\mathbf{v}(x,y,z)=(0,0,z)$, $S=$ upper hemisphere $x^2+y^2+z^2=1$, $z\ge 0$, outward normal.
Step 1 — Quick check: $\mathrm{div}\,\mathbf{v}=1$ suggests Gauss on a solid; $S$ is only half of a closed surface.
Route A — Gauss on the half-ball
Let $V=\{x^2+y^2+z^2\le 1,\ z\ge 0\}$. Then $\partial V=S\cup D$ where $D$ is the disk $z=0$, $x^2+y^2\le 1$.
On $D$: $\mathbf{v}=(0,0,0)$ and outward normal is $-\mathbf{k}$, so $\mathbf{v}\cdot\mathbf{n}=0$ — no flux through the disk.
Route B — Spherical parametrization (§11.3.3, same patch as §11.5 Example A):
On $S$: $z=\cos\vartheta$. With outward orientation, $\mathbf{v}\cdot(\boldsymbol{\Phi}_\vartheta\times\boldsymbol{\Phi}_\varphi)=\cos\vartheta\sin\vartheta\cos^2\vartheta$ (standard unit-sphere patch).
Substitute $u=\cos\vartheta$, $du=-\sin\vartheta\,d\vartheta$:
Check: Routes A and B agree. Contrast §11.5 Example B ($\mathbf{K}=(0,0,z)$ on a graph $z=x^2+y^2$ — same symbol $z$, different surface.
13.7 Exam checklist #
- Closed curve in $\mathbb{R}^2$? → Green or $P_y=Q_x$ / potential (§11.4, Ch 12).
- Area of plane region? → Green with $\mathbf{v}$ such that $(\mathrm{rot}\,\mathbf{v})_z=1$ (§13.4b).
- Closed surface of a solid? → $\mathrm{div}\,\mathbf{v}$ + Gauss (§11.8).
- Flux through one face only? → Gauss on full $V$, subtract other faces (§13.3, §13.6).
- Surface with boundary + circulation? → $\mathrm{rot}\,\mathbf{F}$ + Stokes (§13.4).
- Open surface, flux only? → Kochrezept (§13.5) or Gauss trick from (4).
- Compute $\mathrm{rot}$ or $\mathrm{div}$ first — wrong operator $\Rightarrow$ wrong theorem.
- Orientation: outward $\mathbf{n}$ for Gauss; right-hand rule for Stokes/Green (§13.3).
13.9 Exercises and answers #
A1 Compute $\mathrm{rot}\,(x,y,z)$.
A2 Compute $\mathrm{rot}\,(y,-x,0)$.
B1 Verify $\mathrm{div}(\mathrm{rot}\,\mathbf{v})=0$ for $\mathbf{v}\in C^2$.
B2 On the unit circle in the $xy$-plane, evaluate $\displaystyle\oint(-y\,dx+x\,dy)$ using Stokes on the unit disk.
C1 Flux of $\mathbf{v}=(0,0,z)$ through upper hemisphere $x^2+y^2+z^2=1$, $z\ge 0$ (outward).
C2 Area of $\{x^2+y^2\le 4\}$ using Green with $\mathbf{v}=(0,x,0)$.
C3 Simplify $\mathrm{div}\bigl(x\,\mathrm{rot}(y,-x,0)\bigr)$ at a point (no integral).
Answers
- A1: $\mathrm{rot}\,(x,y,z)=\mathbf{0}$.
- A2: $(0,0,-2)$.
- B1: Write $\mathrm{rot}\,\mathbf{v}=(R_1,R_2,R_3)$; then $\mathrm{div}(\mathrm{rot}\,\mathbf{v})=\partial_x R_1+\partial_y R_2+\partial_z R_3=0$ (equality of mixed partials / Schwarz).
- B2: $2\pi$ (§13.4).
- C1: $\dfrac{2\pi}{3}$ (§13.6).
- C2: $4\pi$ (disk radius $2$: $\oint x\,dy=\int_0^{2\pi}4\cos^2 t\,dt=4\pi$, or $\pi r^2=4\pi$).
- C3: $\mathrm{rot}(y,-x,0)=(0,0,-2)$; $\mathrm{div}(x\,\mathrm{rot}\,\mathbf{K})=\nabla x\cdot\mathrm{rot}\,\mathbf{K}=1\cdot(-2)=-2$ (constant).
Chapter 14 — Topology (Topologie) #
Open sets, closure, compactness, and simply connected domains — why integration theorems have hypotheses on $\Omega$.
Story of the chapter: analysis is not only formulas. Theorems say “$\Omega$ open”, “$K$ compact”, “simply connected” because the shape of the domain matters. This chapter names those shapes and links them to Ch 11 (integrals), Ch 12 (potentials / holes), and Ch 4 (extrema on compact sets).
Prerequisites: Chapter 4 (continuity); helpful after §12.3 (hole vs no hole).
Where to start
Confused by “simply connected” or annulus examples? Read §14.6 first, then return to Ch 12 §12.3.
Setting up a domain for Green / Fubini? §14.1–§14.2 + §14.7.
This chapter is short on purpose: definitions, one table, a few examples — not a general topology course.
14.0 Why topology for analysis #
| Hypothesis you see | What it means (here) | Where it matters |
|---|---|---|
| $\Omega$ open | no boundary points inside $\Omega$; local balls stay in $\Omega$ | domain of $f$, $C^1$ graphs (Ch 8) |
| $K$ compact in $\mathbb{R}^n$ | closed + bounded (Heine–Borel) | EVT: min/max attained (§4.1.6, §8.2.8) |
| $\Omega$ Jordan measurable | bounded; boundary has measure $0$ | Riemann / multiple integrals (§11.1.1) |
| $\Omega$ simply connected | no holes; loops shrink in $\Omega$ | $P_y=Q_x$ $\Rightarrow$ global potential (Ch 12, §14.6) |
| Annulus (hole) | connected but not simply connected | $P_y=Q_x$ but no global $\phi$ (§12.3 Example B) |
14.1 Open and closed sets #
For $x_0\in\mathbb{R}^d$ and $r>0$, the open ball is
Definitions
- $x_0$ is an inner point of $\Omega\subset\mathbb{R}^d$ if $\exists r>0$ with $B_r(x_0)\subseteq\Omega$.
- $\Omega$ is open if every point of $\Omega$ is an inner point.
- $A\subset\mathbb{R}^d$ is closed if $A^c=\mathbb{R}^d\setminus A$ is open.
Useful rules (ZSF 14.1.3–14.1.5)
- $\emptyset$ and $\mathbb{R}^d$ are open (and closed).
- Finite intersections of open sets are open; arbitrary unions of open sets are open.
- Finite unions of closed sets are closed; arbitrary intersections of closed sets are closed.
Sequential test: $A$ is closed iff whenever $x_k\in A$ and $x_k\to x$, then $x\in A$.
Example (classify subsets of $\mathbb{R}$)
(i) $(0,1)$ is open: for $x\in(0,1)$ choose $r=\min\{x,\,1-x\}>0$; then $B_r(x)\subseteq(0,1)$.
(ii) $[0,1]$ is closed: complement $(-\infty,0)\cup(1,\infty)$ is a union of open intervals.
(iii) $[0,1)$ is neither: $0$ is not an inner point (no ball around $0$ lies in $[0,1)$); complement is not open either — common exam trap.
14.2 Interior, closure, boundary #
For $A\subset\mathbb{R}^d$:
- Interior $\mathring{A}$ = union of all open sets contained in $A$ (largest open subset).
- Closure $\overline{A}$ = smallest closed set containing $A$; equivalently $\overline{A}=\{x:\ \exists (x_k)\subset A,\ x_k\to x\}$.
- Boundary $\partial A=\overline{A}\setminus\mathring{A}$.
Relations: $\mathring{A}\subseteq A\subseteq\overline{A}$ and $\overline{A}=\mathring{A}\sqcup\partial A$ (disjoint union).
Boundary point: $x\in\partial A$ iff every ball $B_r(x)$ meets both $A$ and $A^c$.
ZSF table — closure, interior, boundary (§14.5)
| Set $A$ | $\overline{A}$ | $\mathring{A}$ | $\partial A$ |
|---|---|---|---|
| $[0,1]$ | $[0,1]$ | $(0,1)$ | $\{0,1\}$ |
| $(0,1)$ | $[0,1]$ | $(0,1)$ | $\{0,1\}$ |
| $[-1,0)\cup(0,1]$ | $[-1,1]$ | $(-1,0)\cup(0,1)$ | $\{-1,0,1\}$ |
| $\{0\}$ | $\{0\}$ | $\emptyset$ | $\{0\}$ |
| $\mathbb{Q}$ | $\mathbb{R}$ | $\emptyset$ | $\mathbb{R}$ |
| $[0,\infty)$ | $[0,\infty)$ | $(0,\infty)$ | $\{0\}$ |
| $\mathbb{N}$ | $\mathbb{N}\cup\{0\}$ | $\emptyset$ | $\mathbb{N}\cup\{0\}$ |
Example (one row by hand — $[0,1]$)
Interior: every point strictly between $0$ and $1$ has a small interval in $[0,1]$; endpoints are not inner → $\mathring{[0,1]}=(0,1)$.
Closure: limits of sequences in $[0,1]$ cannot escape $[0,1]$ → $\overline{[0,1]}=[0,1]$.
Boundary: $\partial[0,1]=[0,1]\setminus(0,1)=\{0,1\}$.
Check: matches table row; $\mathbb{Q}$ shows $\overline{\mathbb{Q}}=\mathbb{R}$ (rationals dense).
14.3 Compactness in $\mathbb{R}^n$ #
Bounded: $K\subset\mathbb{R}^d$ is bounded if $K\subseteq B_R(0)$ for some $R$.
Compact (Heine–Borel in $\mathbb{R}^d$): for $K\subset\mathbb{R}^d$,
Sequential characterization: $K$ compact $\Leftrightarrow$ every sequence in $K$ has a convergent subsequence with limit in $K$.
Why exams care
- Extreme value theorem: continuous $f$ on compact $K$ attains $\min$ and $\max$ on $K$ (§4.1.6).
- Continuous image: if $f$ continuous and $K$ compact, then $f(K)$ is compact.
- Uniform limits: on compact $K$, uniform limit of continuous functions is continuous (Ch 5).
Example (closed bounded triangle — ZSF 14.4)
Given: $D=\{(x,y)\in\mathbb{R}^2:\ x\ge 0,\ y\ge 0,\ 3x+y\le 3\}$.
Step 1 — Closed (sequence test): let $(x_n,y_n)\in D$ and $(x_n,y_n)\to(x,y)$. Then $x_n\to x$, $y_n\to y$, and $3x_n+y_n\to 3x+y$. Each $x_n\ge 0$, $y_n\ge 0$, $3x_n+y_n\le 3$ pass to the limit → $(x,y)\in D$.
Step 2 — Bounded: from $x\ge 0$, $3x+y\le 3$ get $x\le 1$; similarly $y\le 3$. So $D\subseteq[0,1]\times[0,3]$.
Step 3 — Conclusion: $D$ is closed and bounded in $\mathbb{R}^2$ → compact (Heine–Borel).
Payoff: a continuous function on $\bar D$ has min/max on $D$; useful in existence proofs (Ch 9).
14.4 Continuity and open sets #
For $f:\Omega\to\mathbb{R}^m$ with $\Omega\subset\mathbb{R}^d$ open, the following are equivalent (ZSF 14.3):
- $f$ is continuous (sequential / $\varepsilon$–$\delta$ — Ch 4).
- For every open $V\subset\mathbb{R}^m$, the preimage $f^{-1}(V)=\{x\in\Omega:\ f(x)\in V\}$ is open in $\Omega$.
- For every closed $B\subset\mathbb{R}^m$, $f^{-1}(B)$ is closed in $\Omega$.
Exam use: prove a set is closed by writing it as $f^{-1}(B)$ for continuous $f$ and closed $B$ (as in §14.3 Step 1 with $g_1(x,y)=x$, $g_3(x,y)=3x+y$ and $B=(-\infty,0]$ or $[3,\infty)$).
Norms (one line): equivalent norms on $\mathbb{R}^d$ define the same open sets (ZSF 14.2.2) — so “open” does not depend on whether you use the Euclidean norm or $\|\cdot\|_\infty$.
14.5 Relatively open and closed (ZSF 14.1.10) #
Let $X\subset\mathbb{R}^n$. A set $A\subset X$ is relatively open in $X$ if $A=B\cap X$ for some open $B\subset\mathbb{R}^n$. Similarly relatively closed in $X$.
Example: $X=[0,1)$ and $A=[0,\tfrac12)$. Then $A=(-\tfrac12,\tfrac12)\cap X$ with $(-\tfrac12,\tfrac12)$ open in $\mathbb{R}$, so $A$ is relatively open in $X$ even though $A$ is not open in $\mathbb{R}$.
Why it appears: continuity on a domain $X$ that is not open in $\mathbb{R}^n$ (e.g. a curve or a square with boundary) uses relative topology.
14.6 Connected and simply connected #
Connected: $U\subset\mathbb{R}^d$ cannot be split into two disjoint non-empty sets that are both relatively open in $U$.
Simply connected (exam level in $\mathbb{R}^2$): every closed curve in $U$ can be shrunk continuously to a point without leaving $U$ (no holes). Same hypothesis in Cauchy’s theorem (§16.3).
| Domain | Connected? | Simply connected? | Potential story (Ch 12) |
|---|---|---|---|
| $\mathbb{R}^2$ | yes | yes | $P_y=Q_x$ $\Rightarrow$ global $\phi$ |
| Annulus $\{0.5\le r\le 1.5\}$ | yes | no | $P_y=Q_x$ possible but no global $\phi$ (§12.3 Example B) |
| Two disjoint disks | no | — | path integrals need which component |
Link to conservative fields
On simply connected $\Omega$, for $\mathbf{v}=(P,Q)$: $\mathbf{v}=\nabla\phi$ $\Leftrightarrow$ $P_y=Q_x$ $\Leftrightarrow$ all closed $\oint=0$ (§12.0, §10.1.9).
Hole: $\Omega$ may be connected but not simply connected — then $P_y=Q_x$ is not enough.
14.7 Jordan measurable (link Ch 11) #
Definition (§11.1.1): bounded $\Omega\subset\mathbb{R}^n$ is Jordan measurable iff its boundary $\partial\Omega$ has $(n-1)$-dimensional measure zero.
Intuition: you can approximate $\Omega$ by unions of boxes from inside and outside; “nice” regions for Riemann integration in $\mathbb{R}^n$.
Examples
- Disk, rectangle, ball: Jordan measurable (smooth boundary).
- $\mathbb{Q}\cap[0,1]^2$: bounded but not Jordan measurable (boundary too thick).
Do not confuse: Jordan measurable $\not=$ simply connected; Jordan measurable $\not=$ $C^1$ domain (§11.2, §11.4).
14.8 Exam checklist #
- Open or closed? Inner points / complement open / sequence test (§14.1).
- Interior, closure, boundary? Use definitions or ZSF table (§14.2).
- Compact in $\mathbb{R}^n$? Closed + bounded (§14.3).
- Prove closed: preimage of closed set under continuous map (§14.4).
- Potential / Green: simply connected or hole? (§14.6, Ch 12).
- Set up integral: Jordan measurable? (§14.7, Ch 11).
14.9 Exercises and answers #
A1 Is $(0,1)$ open in $\mathbb{R}$? Is $[0,1]$ closed?
A2 For $A=[0,1]$, find $\mathring{A}$, $\overline{A}$, $\partial A$.
B1 Give a subset of $\mathbb{R}$ that is bounded but not compact.
B2 Is $D=\{(x,y):x\ge 0,\ y\ge 0,\ 3x+y\le 3\}$ compact in $\mathbb{R}^2$?
C1 Is the annulus $\{0.5\le x^2+y^2\le 1.5\}$ simply connected?
C2 Is $\mathbb{Q}\cap[0,1]$ Jordan measurable?
Answers
- A1: $(0,1)$ is open; $[0,1]$ is closed (§14.1).
- A2: $\mathring{A}=(0,1)$, $\overline{A}=[0,1]$, $\partial A=\{0,1\}$ (§14.2).
- B1: $(0,1)$ — bounded in $\mathbb{R}$ but not closed, hence not compact.
- B2: Yes — closed and bounded (§14.3).
- C1: No — a loop around the inner circle cannot shrink to a point inside the annulus (§14.6, §12.3).
- C2: No — boundary too large / not a “nice” Jordan region (§14.7).
Chapter 15 — Reference formulas #
Compact reference for exam lookup — standard identities, roots, and $dV$ factors in one place.
Use when: during a calculation you need a formula you do not remember.
Quick navigation
| You need | Section |
|---|---|
| Quadratic roots | §15.1–15.2 |
| $\sin/\cos$ at special angles | §15.3 |
| Addition / power trig | §15.4 |
| $\cosh/\sinh$, $\operatorname{arsinh}$ | §15.5 |
| $t=\tan(x/2)$ substitution | §15.6 |
| Matrix inverse / definiteness | §15.7 |
| Diagonalize $A$ | §15.8 |
| Binomial / factor $x^n-1$ | §15.9 |
| Volume / surface of standard body | §15.11 |
| Polar / cylindrical / spherical $dV$ | §15.12 |
| Standard antiderivatives | §15.13 |
15.1 Mitternachtsformel #
For $a\neq 0$:
15.2 $pq$-formula #
For $x^2+px+q=0$ (leading coefficient $1$):
15.3 Trigonometric values #
| $\varphi$ | $0$ | $\pi/6$ | $\pi/4$ | $\pi/3$ | $\pi/2$ | $2\pi/3$ | $3\pi/4$ | $5\pi/6$ | $\pi$ |
|---|---|---|---|---|---|---|---|---|---|
| $\sin\varphi$ | $0$ | $\tfrac12$ | $\tfrac{\sqrt2}{2}$ | $\tfrac{\sqrt3}{2}$ | $1$ | $\tfrac{\sqrt3}{2}$ | $\tfrac{\sqrt2}{2}$ | $\tfrac12$ | $0$ |
| $\cos\varphi$ | $1$ | $\tfrac{\sqrt3}{2}$ | $\tfrac{\sqrt2}{2}$ | $\tfrac12$ | $0$ | $-\tfrac12$ | $-\tfrac{\sqrt2}{2}$ | $-\tfrac{\sqrt3}{2}$ | $-1$ |
| $\tan\varphi$ | $0$ | $\tfrac{1}{\sqrt3}$ | $1$ | $\sqrt3$ | — | $-\sqrt3$ | $-1$ | $-\tfrac{1}{\sqrt3}$ | $0$ |
15.4 Addition theorems and powers #
Core
Product-to-sum
Inverse trig (domain restrictions as in course)
Power reductions
Phase shifts
15.5 Hyperbolic identities #
Same bridge as Euler’s formula in §16.1.
Inverses
15.6 Tangent half-angle #
With $t=\tan\frac{x}{2}$:
15.7 Matrices — inverse and definiteness #
$2\times 2$ inverse ($\det A=ad-bc\neq 0$):
Definiteness (symmetric $H$)
| Method | Rule |
|---|---|
| I — Eigenvalues | all $\lambda_i>0$ $\Rightarrow$ positive definite; all $<0$ $\Rightarrow$ negative definite; mixed signs $\Rightarrow$ indefinite |
| II — Sylvester ($n\times n$ symmetric) | all leading principal minors $A_1,A_2,\ldots,A_n>0$ $\Rightarrow$ positive definite; alternating signs $A_1<0,A_2>0,\ldots$ $\Rightarrow$ negative definite |
15.8 Diagonalization recipe #
For $A\in\mathbb{R}^{n\times n}$:
- Characteristic polynomial: $\det(A-\lambda I)=0$ $\Rightarrow$ eigenvalues $\gamma_1,\ldots,\gamma_n$.
- For each $\gamma_k$, solve $(A-\gamma_k I)\mathbf{v}=\mathbf{0}$ for eigenvector $\mathbf{v}_k$.
- $S=[\mathbf{v}_1\ \cdots\ \mathbf{v}_n]$, $D=\mathrm{diag}(\gamma_1,\ldots,\gamma_n)$.
- If $S$ invertible: $A=SDS^{-1}$.
15.9 Binomial and polynomial tricks #
Factoring
15.10 $\max$ as formula #
(Combination of continuous functions $\Rightarrow$ continuous.)
15.11 Geometric sets #
| Shape | Set / equation | Area / length / volume / surface |
|---|---|---|
| Circle (radius $r$) | $(x-x_0)^2+(y-y_0)^2=r^2$ | $A=\pi r^2$, $U=2\pi r$ |
| Sphere (radius $r$) | $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2$ | $V=\frac{4}{3}\pi r^3$, $S=4\pi r^2$ |
| Cylinder | $(x-x_0)^2+(y-y_0)^2=r^2$, $0\le z\le h$ | $V=\pi r^2 h$, mantle $M=2\pi rh$, $S=M+2\pi r^2$ |
| Cone | $x^2+y^2=\frac{r^2}{h^2}(h-z)^2$, $0\le z\le h$ | $V=\frac{1}{3}\pi r^2 h$ |
| Ellipse | $\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1$ | boundary: $(a\cos t,b\sin t)$ |
| Ellipsoid | $\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}+\frac{(z-z_0)^2}{c^2}=1$ | volume via substitution |
| Elliptic paraboloid | $\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=z-z_0$ | graph $z=z_0+\cdots$ |
| Torus (major $R$, tube $r$, $r<R$) | $(\sqrt{x^2+y^2}-R)^2+z^2=r^2$ | $V=2\pi^2 R r^2$ |
15.12 Coordinate parametrizations #
| Coordinates | $\Phi$ | $\lvert\det D\Phi\rvert$ / element |
|---|---|---|
| Polar $(r,\varphi)$ | $(r\cos\varphi,r\sin\varphi)$ | $dA=r\,dr\,d\varphi$ |
| Cylindrical $(r,\varphi,z)$ | $(r\cos\varphi,r\sin\varphi,z)$ | $dV=r\,dr\,d\varphi\,dz$ |
| Spherical $(r,\vartheta,\varphi)$ | $(r\sin\vartheta\cos\varphi,\,r\sin\vartheta\sin\varphi,\,r\cos\vartheta)$ | $dV=r^2\sin\vartheta\,dr\,d\vartheta\,d\varphi$ |
Graph surface $z=f(x,y)$ over $D$:
15.13 Quick integral primitives #
Standard antiderivatives:
| Integral | Result |
|---|---|
| $\displaystyle\int\frac{dx}{\sqrt{1-x^2}}$ | $\arcsin x+C$ |
| $\displaystyle\int\frac{dx}{1+x^2}$ | $\arctan x+C$ |
| $\displaystyle\int\frac{dx}{\sqrt{x^2+1}}$ | $\operatorname{arsinh} x+C$ |
| $\displaystyle\int\frac{dx}{\sqrt{x^2-1}}$ | $\operatorname{arcosh} x+C$ $(x>1)$ |
| $\displaystyle\int\tan x\,dx$ | $-\ln\lvert\cos x\rvert+C$ |
| $\displaystyle\int\ln x\,dx$ | $x(\ln\lvert x\rvert-1)+C$ |
| $\displaystyle\int e^{ax}\,dx$ | $\frac{1}{a}e^{ax}+C$ |
| $\displaystyle\int x^n e^{ax}\,dx$ | integration by parts |
15.14 Quick lookup exercises #
A1 Solve $x^2-5x+6=0$ via $pq$. A2 Exact value of $\sin(75°)$ using $\sin(45°+30°)$.
B1 Is $\begin{pmatrix}2&1\\1&2\end{pmatrix}$ positive definite? B2 Volume of a cone radius $3$, height $4$.
Answers
- A1: $p=-5$, $q=6$ $\Rightarrow$ $x=2$ or $x=3$.
- A2: $\sin 75°=\sin 45°\cos 30°+\cos 45°\sin 30°=\frac{\sqrt6+\sqrt2}{4}$.
- B1: Yes — leading minors $2>0$, $\det=3>0$ (§15.7).
- B2: $V=\frac{1}{3}\pi\cdot 9\cdot 4=12\pi$ (§15.11).
Chapter 16 — Complex numbers ($\mathbb{C}$) #
Algebra, polar form, holomorphic functions, and contour integrals at exam level.
Story of the chapter: $\mathbb{C}$ completes the algebra of polynomials and encodes rotation and oscillation through $e^{i\theta}$. For analysis exams you need three layers: arithmetic/polar, holomorphic + Cauchy–Riemann, and contour integrals / residues.
Prerequisites: Chapter 2 (limits), Chapter 6, Chapter 8 (partials for CR).
16.1 Algebra, polar form, and Euler #
Basics
$z=x+iy$, $\overline{z}=x-iy$, $\Re z=x$, $\Im z=y$, $i^2=-1$.
Division
In polar form: divide moduli, subtract arguments.
Polar / Euler
Multiplication: $|z_1 z_2|=|z_1||z_2|$, $\arg(z_1 z_2)=\arg z_1+\arg z_2$ mod $2\pi$.
De Moivre
$n$th roots of $z\neq 0$: there are exactly $n$ distinct roots
Link: complex characteristic roots $\alpha\pm i\beta$ in §7.2 give $e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)$ — the same rotation as $e^{(\alpha+i\beta)x}$.
Typical mistakes
- Treating $\arg(z_1 z_2)=\arg z_1+\arg z_2$ as an exact equality without mod $2\pi$.
- Forgetting all $n$ roots of $z^n=w$ (e.g. $z^2=-1$ gives $\pm i$).
16.2 Holomorphic functions and Cauchy–Riemann #
$f:\Omega\subset\mathbb{C}\to\mathbb{C}$, write $f=u+iv$ with $u,v:\Omega\to\mathbb{R}$.
Complex derivative at $z_0$:
(limit must exist for every complex $h\to 0$, not only along axes).
Holomorphic on $\Omega$ if $f'$ exists at every point of $\Omega$.
Cauchy–Riemann (CR): if $u,v\in C^1$, then $f$ holomorphic on $\Omega$ $\Rightarrow$
If $u,v\in C^1$ and CR hold on $\Omega$, then $f$ is holomorphic on $\Omega$.
Harmonic: if $f$ holomorphic, then $u$ and $v$ are harmonic ($\Delta u=0$, $\Delta v=0$).
Examples
- $f(z)=z^2=(x^2-y^2)+2ixy$: $u_x=2x=v_y$, $u_y=-2y=-v_x$ — holomorphic.
- $f(x,y)=x$ only (no $y$ in imaginary part): $u_x=1$, $v_y=0$ — not holomorphic.
Typical mistake Checking $u_x=v_y$ but forgetting $u_y=-v_x$ (or testing CR only along $x$ / $y$ directions).
16.3 Cauchy theorems and contour setup #
Contour $\gamma:[a,b]\to\mathbb{C}$ piecewise $C^1$:
Cauchy (simply connected): if $f$ is holomorphic on a simply connected open $\Omega$ and $\gamma$ is a closed contour in $\Omega$, then
Hypothesis: “simply connected” means no holes — see §14.6. On an annulus, $f$ can be holomorphic with $\oint\neq 0$.
Path independence: on simply connected $\Omega$, $\int_\gamma f\,dz$ depends only on the endpoints of $\gamma$.
Typical mistakes
- Applying Cauchy on a domain with a hole without checking simply connected.
- Parametrizing $\gamma(t)$ but omitting $\gamma'(t)$ in $dz$.
16.4 Cauchy integral formula #
If $f$ is holomorphic on $|z-z_0|<R$ and $|w-z_0|<r<R$:
Derivatives ($n\ge 0$):
Use when: the integrand is $\dfrac{g(z)}{(z-a)^{n+1}}$ with $g$ holomorphic inside the circle — read off $f^{(n)}(a)$ from $g$.
16.5 Residues and contour integrals #
Residue at an isolated singularity $a$:
Simple pole at $a$:
Standard templates
- $\operatorname{Res}\bigl(\dfrac{1}{z-a},a\bigr)=1$.
- For $\dfrac{g(z)}{z-a}$ with $g$ holomorphic and $g(a)\neq 0$: $\operatorname{Res}(f,a)=g(a)$.
Pole of order $m\ge 2$ at $a$:
Residue theorem
Contour Kochrezept — exam card
- Locate poles inside $\gamma$ (ignore poles outside).
- Compute residues (simple-pole rule or order-$m$ formula).
- Sum residues and multiply by $2\pi i$.
Typical mistakes
- Counting a pole outside $\gamma$.
- Using the simple-pole formula on a double pole.
Out of scope here: full Laurent-series chapter; essential singularities (only: do not blindly use simple-pole rules).
Example A — $\oint_{|z|=2}\dfrac{1}{z^2+1}\,dz$
Factor $z^2+1=(z-i)(z+i)$. Poles $z=\pm i$ lie inside $|z|=2$.
At $z=i$: $\operatorname{Res}=\displaystyle\lim_{z\to i}\frac{z-i}{z^2+1}=\frac{1}{2i}$.
At $z=-i$: $\operatorname{Res}=\displaystyle\lim_{z\to -i}\frac{z+i}{z^2+1}=\frac{1}{-2i}=-\frac{1}{2i}$.
Sum $=0$, so $\oint_{|z|=2}\dfrac{1}{z^2+1}\,dz=0$.
Example B — $\oint_{|z|=1}\dfrac{e^z}{z^2}\,dz$
Pole at $z=0$ of order $2$. With $g(z)=e^z$ holomorphic,
Hence $\oint_{|z|=1}\dfrac{e^z}{z^2}\,dz=2\pi i$.
Example C — $\displaystyle\int_{-\infty}^{\infty}\frac{1}{1+x^2}\,dx=\pi$
Close the real axis by the upper semicircle $\gamma_R$ of radius $R$. For $R\to\infty$, $\displaystyle\int_{\gamma_R}\frac{1}{1+z^2}\,dz\to 0$ (Jordan / estimate: $|1/(1+z^2)|\le C/R^2$ on the arc, length $\pi R$).
Only pole inside: $z=i$, with $\operatorname{Res}=\dfrac{1}{2i}$.
16.9 Exercises #
A1 Compute $(1+i)^8$ using De Moivre.
A2 Verify CR for $f(z)=z^2$ written as $u(x,y)+iv(x,y)$.
A3 $\displaystyle\oint_{|z|=3}\frac{1}{z(z-2)}\,dz$.
B1 Polar form of $-1+i\sqrt{3}$.
B2 Is $e^z$ holomorphic on $\mathbb{C}$? (Use $e^{x+iy}=e^x(\cos y+i\sin y)$.)
B3 $\displaystyle\oint_{|z|=2}\frac{z}{z^2+1}\,dz$.
Answers
- A1: $|1+i|=\sqrt2$, $\arg(1+i)=\pi/4$ $\Rightarrow$ $(1+i)^8=(\sqrt2)^8 e^{i\cdot 8\cdot\pi/4}=16 e^{i2\pi}=16$.
- A2: $u=x^2-y^2$, $v=2xy$; $u_x=2x=v_y$, $u_y=-2y=-v_x$ — holomorphic.
- A3: Partial fractions $\dfrac{1}{z(z-2)}=\dfrac{1}{2(z-2)}-\dfrac{1}{2z}$. Residues $+\tfrac12$ at $z=2$ and $-\tfrac12$ at $z=0$; both inside $|z|=3$; sum $=0$.
- B1: $|z|=2$, $\arg z=2\pi/3$ $\Rightarrow$ $z=2e^{i2\pi/3}$.
- B2: Yes — $e^z$ is entire (holomorphic on all of $\mathbb{C}$).
- B3: Poles at $\pm i$ inside $|z|=2$. At $z=i$: $\operatorname{Res}=\dfrac{i}{2i}=\tfrac12$; at $z=-i$: $\dfrac{-i}{-2i}=\tfrac12$; integral $=2\pi i$.
Chapter 17 — Additional integrals (Zusätzliche Integrale) #
Supplement to Chapter 10 — special forms from the ZSF / Dirren appendix.
Use when: the integrand matches one of the templates below and §10.2–10.3 did not immediately apply.
17.1 Substitution templates #
| Form | Substitution |
|---|---|
| $\int f(e^x,\sinh x,\cosh x)\,dx$ | $u=e^x$ or hyperbolic $u$ |
| $\int f(x,\sqrt{1-x^2})\,dx$ | $x=\sin u$ |
| $\int f(x,\sqrt{1+x^2})\,dx$ | $x=\sinh u$ |
| $\int f(x,\sqrt{x^2-1})\,dx$ | $x=\cosh u$ |
| $\int \dfrac{dx}{\sqrt{a^2-x^2}}$ | $x=a\sin u$ → $\arcsin$ |
| $\int \dfrac{dx}{1+x^2}$ | $\arctan$ |
| $\int \dfrac{dx}{\sqrt{x^2+a^2}}$ | $\operatorname{arsinh}$ after $x=a\sinh u$ |
17.2 Worked example #
(see §10.3 rationalization family).
Chapter 18 — Fourier series (Fourierreihe) #
Orthogonal expansion of periodic functions — full theory beyond §10.3.6.
Story of the chapter: periodic signals repeat; Fourier series splits them into sines and cosines (eigenmodes of translation). Coefficients are projections using orthogonality.
Prerequisites: Chapter 3, §10.3.6.
18.1 Orthogonality on $[-\pi,\pi]$ #
18.2 Fourier coefficients #
For $2\pi$-periodic $f$ integrable on $[-\pi,\pi]$:
Even $f$: only cosines; odd $f$: only sines.
18.3 Convergence (exam level) #
For piecewise $C^1$ periodic $f$, series converges to $\tfrac12(f(x^+)+f(x^-))$ at jumps.
Typical mistake Differentiating termwise without checking uniform convergence of derivative series.
18.4 Parseval (energy) #
18.9 Exercises #
A1 Fourier coefficients of $f(x)=x$ on $(-\pi,\pi)$ (odd → sine series).
B1 Square wave $f(x)=\mathrm{sgn}(\sin x)$ — sketch partial sums.